RS Aggarwal Solutions for Class 12 Chapter 25 Product of Three Vectors

Question 1. Prove that (i) \( [\hat{i} \quad \hat{j} \quad \hat{k}] = [\hat{j} \quad \hat{k} \quad \hat{i}] = [\hat{k} \quad \hat{i} \quad \hat{j}] = 1 \) (ii) \( [\hat{i} \quad \hat{k} \quad \hat{j}] = [\hat{k} \quad \hat{j} \quad \hat{i}] = [\hat{j} \quad \hat{i} \quad \hat{k}] = -1 \)
Answer: Let \( \hat{i}, \hat{j}, \hat{k} \) denote unit vectors in the direction of positive X-axis, Y-axis, and Z-axis respectively. The magnitudes satisfy \( |\hat{i}| = |\hat{j}| = |\hat{k}| = 1 \).

Part (i):
\( [\hat{i} \quad \hat{j} \quad \hat{k}] = \hat{i} \cdot (\hat{j} \times \hat{k}) \)

Since \( \hat{j} \times \hat{k} = \hat{i} \):
\( = \hat{i} \cdot \hat{i} = 1 \)

Therefore, \( [\hat{i} \quad \hat{j} \quad \hat{k}] = 1 \) ... (eq 1)

\( [\hat{j} \quad \hat{k} \quad \hat{i}] = \hat{j} \cdot (\hat{k} \times \hat{i}) \)

Since \( \hat{k} \times \hat{i} = \hat{j} \):
\( = \hat{j} \cdot \hat{j} = 1 \)

Therefore, \( [\hat{j} \quad \hat{k} \quad \hat{i}] = 1 \) ... (eq 2)

\( [\hat{k} \quad \hat{i} \quad \hat{j}] = \hat{k} \cdot (\hat{i} \times \hat{j}) \)

Since \( \hat{i} \times \hat{j} = \hat{k} \):
\( = \hat{k} \cdot \hat{k} = 1 \)

Therefore, \( [\hat{k} \quad \hat{i} \quad \hat{j}] = 1 \) ... (eq 3)

From equations (1), (2), and (3): \( [\hat{i} \quad \hat{j} \quad \hat{k}] = [\hat{j} \quad \hat{k} \quad \hat{i}] = [\hat{k} \quad \hat{i} \quad \hat{j}] = 1 \)

Part (ii):
\( [\hat{i} \quad \hat{k} \quad \hat{j}] = \hat{i} \cdot (\hat{k} \times \hat{j}) \)

Since \( \hat{k} \times \hat{j} = -\hat{i} \):
\( = \hat{i} \cdot (-\hat{i}) = -1 \)

Therefore, \( [\hat{i} \quad \hat{k} \quad \hat{j}] = -1 \) ... (eq 1)

\( [\hat{k} \quad \hat{j} \quad \hat{i}] = \hat{k} \cdot (\hat{j} \times \hat{i}) \)

Since \( \hat{j} \times \hat{i} = -\hat{k} \):
\( = \hat{k} \cdot (-\hat{k}) = -1 \)

Therefore, \( [\hat{k} \quad \hat{j} \quad \hat{i}] = -1 \) ... (eq 2)

\( [\hat{j} \quad \hat{i} \quad \hat{k}] = \hat{j} \cdot (\hat{i} \times \hat{k}) \)

Since \( \hat{i} \times \hat{k} = -\hat{j} \):
\( = \hat{j} \cdot (-\hat{j}) = -1 \)

Therefore, \( [\hat{j} \quad \hat{i} \quad \hat{k}] = -1 \) ... (eq 3)

From equations (1), (2), and (3): \( [\hat{i} \quad \hat{k} \quad \hat{j}] = [\hat{k} \quad \hat{j} \quad \hat{i}] = [\hat{j} \quad \hat{i} \quad \hat{k}] = -1 \)

Hence Proved.
In simple words: When you take the scalar triple product of three unit basis vectors in their standard order (i, j, k going around a circle), you always get 1. When you reverse the order (going backward around the circle), you always get -1.

Exam Tip: Remember that a cyclic permutation of vectors in a scalar triple product keeps the value the same, while reversing any two vectors changes the sign. This is a key property that appears frequently in higher-level vector problems.

Notes:

• A cyclic change of vectors in a scalar triple product preserves its value. That is, \( [\bar{a} \quad \bar{b} \quad \bar{c}] = [\bar{b} \quad \bar{c} \quad \bar{a}] = [\bar{c} \quad \bar{a} \quad \bar{b}] \)
• The scalar triple product of unit vectors taken in a clockwise direction equals 1, and that of unit vectors taken in an anticlockwise direction equals -1. For instance, \( [\hat{i} \quad \hat{j} \quad \hat{k}] = 1 \) and \( [\hat{k} \quad \hat{j} \quad \hat{i}] = -1 \)

Question 2. Find \( [\bar{a} \quad \bar{b} \quad \bar{c}] \) when (i) \( \bar{a} = 2\hat{i} + \hat{j} + 3\hat{k}, \bar{b} = -\hat{i} + 2\hat{j} + \hat{k} \) and \( \bar{c} = 3\hat{i} + \hat{j} + 2\hat{k} \) (ii) \( \bar{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}, \bar{b} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \bar{c} = 3\hat{i} - \hat{j} + 2\hat{k} \) (iii) \( \bar{a} = 2\hat{i} - 3\hat{j}, \bar{b} = \hat{i} + \hat{j} - \hat{k} \) and \( \bar{c} = 3\hat{i} - \hat{k} \)
Answer:

Part (i): Given vectors:
\( \bar{a} = 2\hat{i} + \hat{j} + 3\hat{k} \)
\( \bar{b} = -\hat{i} + 2\hat{j} + \hat{k} \)
\( \bar{c} = 3\hat{i} + \hat{j} + 2\hat{k} \)

The scalar triple product is calculated using the determinant:
\[ [\bar{a} \quad \bar{b} \quad \bar{c}] = \begin{vmatrix} 2 & 1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & 2 \end{vmatrix} \]

Expanding along the first row:
\( = 2(2 \times 2 - 1 \times 1) - 1((-1) \times 2 - 3 \times 1) + 3((-1) \times 1 - 3 \times 2) \)
\( = 2(3) - 1(-5) + 3(-7) \)
\( = 6 + 5 - 21 \)
\( = -10 \)

Therefore, \( [\bar{a} \quad \bar{b} \quad \bar{c}] = -10 \)

Part (ii): Given vectors:
\( \bar{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \)
\( \bar{b} = \hat{i} + 2\hat{j} - \hat{k} \)
\( \bar{c} = 3\hat{i} - \hat{j} + 2\hat{k} \)

The scalar triple product is:
\[ [\bar{a} \quad \bar{b} \quad \bar{c}] = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \]

Expanding along the first row:
\( = 2(2 \times 2 - (-1) \times (-1)) - (-3)(1 \times 2 - 3 \times (-1)) + 4(1 \times (-1) - 3 \times 2) \)
\( = 2(3) - (-3)(5) + 4(-7) \)
\( = 6 + 15 - 28 \)
\( = -7 \)

Therefore, \( [\bar{a} \quad \bar{b} \quad \bar{c}] = -7 \)

Part (iii): Given vectors:
\( \bar{a} = 2\hat{i} - 3\hat{j} \)
\( \bar{b} = \hat{i} + \hat{j} - \hat{k} \)
\( \bar{c} = 3\hat{i} - \hat{k} \)

Writing in standard form (including zero components):
\( \bar{a} = 2\hat{i} - 3\hat{j} + 0\hat{k} \)
\( \bar{b} = \hat{i} + \hat{j} - \hat{k} \)
\( \bar{c} = 3\hat{i} + 0\hat{j} - \hat{k} \)

The scalar triple product is:
\[ [\bar{a} \quad \bar{b} \quad \bar{c}] = \begin{vmatrix} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{vmatrix} \]

Expanding along the first row:
\( = 2(1 \times (-1) - (-1) \times 0) - (-3)(1 \times (-1) - 3 \times (-1)) + 0(1 \times 0 - 3 \times 1) \)
\( = 2(-1) - (-3)(2) + 0 \)
\( = -2 + 6 \)
\( = 4 \)

Therefore, \( [\bar{a} \quad \bar{b} \quad \bar{c}] = 4 \)
In simple words: To find the scalar triple product, arrange the components of the three vectors as rows in a determinant and evaluate it using cofactor expansion or any standard method for computing 3×3 determinants.

Exam Tip: Always write vectors in complete form with all three components (using 0 if needed). Be careful with signs when expanding the determinant - a common error is forgetting the alternating signs in cofactor expansion.

Question 3. Find the volume of the parallelepiped whose coterminous edges are represented by the vectors (i) \( \bar{a} = \hat{i} + \hat{j} + \hat{k}, \bar{b} = \hat{i} - \hat{j} + \hat{k}, \bar{c} = \hat{i} + 2\hat{j} - \hat{k} \) (ii) \( \bar{a} = -3\hat{i} + 7\hat{j} + 5\hat{k}, \bar{b} = -5\hat{i} + 7\hat{j} - 3\hat{k}, \bar{c} = 7\hat{i} - 5\hat{j} - 3\hat{k} \) (iii) \( \bar{a} = \hat{i} - 2\hat{j} + 3\hat{k}, \bar{b} = 2\hat{i} + \hat{j} - \hat{k}, \bar{c} = \hat{j} + \hat{k} \) (iv) \( \bar{a} = 6\hat{i}, \bar{b} = 2\hat{j}, \bar{c} = 5\hat{k} \)
Answer:

Part (i): Given coterminous edges:
\( \bar{a} = \hat{i} + \hat{j} + \hat{k} \)
\( \bar{b} = \hat{i} - \hat{j} + \hat{k} \)
\( \bar{c} = \hat{i} + 2\hat{j} - \hat{k} \)

Volume of the parallelepiped is given by \( V = |[\bar{a} \quad \bar{b} \quad \bar{c}]| \)

\[ V = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} \]

Expanding along the first row:
\( = 1((-1) \times (-1) - 2 \times 1) - 1(1 \times (-1) - 1 \times 1) + 1(1 \times 2 - 1 \times (-1)) \)
\( = 1(-1) - 1(-2) + 1(3) \)
\( = -1 + 2 + 3 \)
\( = 4 \)

Therefore, Volume of parallelepiped = 4 cubic units

Part (ii): Given coterminous edges:
\( \bar{a} = -3\hat{i} + 7\hat{j} + 5\hat{k} \)
\( \bar{b} = -5\hat{i} + 7\hat{j} - 3\hat{k} \)
\( \bar{c} = 7\hat{i} - 5\hat{j} - 3\hat{k} \)

\[ V = \begin{vmatrix} -3 & 7 & 5 \\ -5 & 7 & -3 \\ 7 & -5 & -3 \end{vmatrix} \]

Expanding along the first row:
\( = -3(7 \times (-3) - (-5) \times (-3)) - 7((-5) \times (-3) - 7 \times (-3)) + 5((-5) \times (-5) - 7 \times 7) \)
\( = -3(-36) - 7(36) + 5(-24) \)
\( = 108 - 252 - 120 \)
\( = -264 \)

Since volume is always positive, \( V = |-264| = 264 \)

Therefore, Volume of parallelepiped = 264 cubic units

Part (iii): Given coterminous edges:
\( \bar{a} = \hat{i} - 2\hat{j} + 3\hat{k} \)
\( \bar{b} = 2\hat{i} + \hat{j} - \hat{k} \)
\( \bar{c} = \hat{j} + \hat{k} \)

Writing in standard form (including zero components):
\( \bar{c} = 0\hat{i} + \hat{j} + \hat{k} \)

\[ V = \begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 0 & 1 & 1 \end{vmatrix} \]

Expanding along the first row:
\( = 1(1 \times 1 - (-1) \times 1) - (-2)(2 \times 1 - 0 \times (-1)) + 3(2 \times 1 - 0 \times 1) \)
\( = 1(2) - (-2)(2) + 3(2) \)
\( = 2 + 4 + 6 \)
\( = 12 \)

Therefore, Volume of parallelepiped = 12 cubic units

Part (iv): Given coterminous edges:
\( \bar{a} = 6\hat{i} \)
\( \bar{b} = 2\hat{j} \)
\( \bar{c} = 5\hat{k} \)

Writing in standard form:
\( \bar{a} = 6\hat{i} + 0\hat{j} + 0\hat{k} \)
\( \bar{b} = 0\hat{i} + 2\hat{j} + 0\hat{k} \)
\( \bar{c} = 0\hat{i} + 0\hat{j} + 5\hat{k} \)

\[ V = \begin{vmatrix} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{vmatrix} \]

This is a diagonal matrix, so the determinant is the product of diagonal elements:
\( V = 6 \times 2 \times 5 = 60 \)

Therefore, Volume of parallelepiped = 60 cubic units
In simple words: The volume of a parallelepiped formed by three vectors equals the absolute value of their scalar triple product. Set up a determinant with the vector components as rows and compute it using standard methods.

Exam Tip: Always take the absolute value of the scalar triple product result since volume is a positive quantity. For vectors aligned with coordinate axes, the volume is simply the product of their magnitudes. Double-check your determinant expansion by using a different row or column if unsure.

Question 4. Show that the vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are coplanar, when
(i) \(\bar{a} = \hat{i} - 2\hat{j} + 3\hat{k}\), \(\bar{b} = -2\hat{i} + 3\hat{j} - 4\hat{k}\) and \(\bar{c} = \hat{i} - 3\hat{j} + 5\hat{k}\)
(ii) \(\bar{a} = \hat{i} + 3\hat{j} + \hat{k}\), \(\bar{b} = 2\hat{i} - \hat{j} - \hat{k}\) and \(\bar{c} = 7\hat{j} + 3\hat{k}\)
(iii) \(\bar{a} = 2\hat{i} - \hat{j} + 2\hat{k}\), \(\bar{b} = \hat{i} + 2\hat{j} - 3\hat{k}\) and \(\bar{c} = 3\hat{i} - 4\hat{j} + 7\hat{k}\)
Answer:
(i) For the given vectors:
\(\bar{a} = \hat{i} - 2\hat{j} + 3\hat{k}\)
\(\bar{b} = -2\hat{i} + 3\hat{j} - 4\hat{k}\)
\(\bar{c} = \hat{i} - 3\hat{j} + 5\hat{k}\)

To demonstrate that these vectors are coplanar, we calculate the scalar triple product. Three vectors are coplanar when their scalar triple product equals zero, that is, \([\bar{a} \, \bar{b} \, \bar{c}] = 0\).

\([\bar{a} \, \bar{b} \, \bar{c}] = \begin{vmatrix} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix}\)

\(= 1(3 \times 5 - (-3) \times (-4)) - (-2)((-2) \times 5 - 1 \times (-4)) + 3((-2) \times (-3) - 1 \times 3)\)

\(= 1(3) + 2(-6) + 3(3)\)
\(= 3 - 12 + 9\)
\(= 0\)

Since the scalar triple product equals zero, the vectors are coplanar.

(ii) For the given vectors:
\(\bar{a} = \hat{i} + 3\hat{j} + \hat{k}\)
\(\bar{b} = 2\hat{i} - \hat{j} - \hat{k}\)
\(\bar{c} = 7\hat{j} + 3\hat{k}\)

\([\bar{a} \, \bar{b} \, \bar{c}] = \begin{vmatrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & 7 & 3 \end{vmatrix}\)

\(= 1((-1) \times 3 - 7 \times (-1)) - 3(2 \times 3 - 0 \times (-1)) + 1(2 \times 7 - 0 \times (-1))\)

\(= 1(4) - 3(6) + 1(14)\)
\(= 4 - 18 + 14\)
\(= 0\)

The scalar triple product is zero, confirming the vectors are coplanar.

(iii) For the given vectors:
\(\bar{a} = 2\hat{i} - \hat{j} + 2\hat{k}\)
\(\bar{b} = \hat{i} + 2\hat{j} - 3\hat{k}\)
\(\bar{c} = 3\hat{i} - 4\hat{j} + 7\hat{k}\)

\([\bar{a} \, \bar{b} \, \bar{c}] = \begin{vmatrix} 2 & -1 & 2 \\ 1 & 2 & -3 \\ 3 & -4 & 7 \end{vmatrix}\)

\(= 2(2 \times 7 - (-4) \times (-3)) - (-1)(1 \times 7 - 3 \times (-3)) + 2(1 \times (-4) - 3 \times 2)\)

\(= 2(4) - (-1)(16) + 2(-10)\)
\(= 4 + 16 - 20\)
\(= 0\)

The scalar triple product equals zero, proving the vectors are coplanar.
In simple words: Three vectors sit in the same plane when the scalar triple product is zero. This happens because the volume of the shape they form becomes zero when they are flat.

Exam Tip: Always compute the determinant carefully, expanding along any row or column. A scalar triple product of zero is the key indicator of coplanarity - verify this condition explicitly in your final answer.

Question 5. Find the value of λ for which the vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are coplanar, when
(i) \(\bar{a} = (2\hat{i} - \hat{j} + \hat{k})\), \(\bar{b} = (\hat{i} + 2\hat{j} + 3\hat{k})\) and \(\bar{c} = (3\hat{i} + \lambda\hat{j} + 5\hat{k})\)
(ii) \(\bar{a} = \lambda\hat{i} - 10\hat{j} - 5\hat{k}\), \(\bar{b} = -7\hat{i} - 5\hat{j}\) and \(\bar{c} = \hat{i} - 4\hat{j} - 3\hat{k}\)
(iii) \(\bar{a} = \hat{i} - \hat{j} + \hat{k}\), \(\bar{b} = 2\hat{i} + \hat{j} - \hat{k}\) and \(\bar{c} = \lambda\hat{i} - \hat{j} + \lambda\hat{k}\)
Answer:
(i) For the given vectors:
\(\bar{a} = 2\hat{i} - \hat{j} + \hat{k}\)
\(\bar{b} = \hat{i} + 2\hat{j} + 3\hat{k}\)
\(\bar{c} = 3\hat{i} + \lambda\hat{j} + 5\hat{k}\)

Since the vectors are coplanar, their scalar triple product must be zero:
\([\bar{a} \, \bar{b} \, \bar{c}] = 0\) .........(1)

\(\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & \lambda & 5 \end{vmatrix} = 0\)

\(= 2(2 \times 5 - 3 \times \lambda) - (-1)(1 \times 5 - 3 \times 3) + 1(1 \times \lambda - 3 \times 2) = 0\)

\(= 2(10 - 3\lambda) + 4 + (\lambda - 6) = 0\)

\(= 20 - 6\lambda + 4 + \lambda - 6 = 0\)

\([\bar{a} \, \bar{b} \, \bar{c}] = 18 - 5\lambda\) .........(2)

From (1) and (2):
\(18 - 5\lambda = 0\)
\(5\lambda = 18\)
\(\lambda = 3.6\)

(ii) For the given vectors:
\(\bar{a} = \lambda\hat{i} - 10\hat{j} - 5\hat{k}\)
\(\bar{b} = -7\hat{i} - 5\hat{j} + 0\hat{k}\)
\(\bar{c} = \hat{i} - 4\hat{j} - 3\hat{k}\)

Since the vectors are coplanar:
\([\bar{a} \, \bar{b} \, \bar{c}] = 0\) .........(1)

\(\begin{vmatrix} \lambda & -10 & -5 \\ -7 & -5 & 0 \\ 1 & -4 & -3 \end{vmatrix} = 0\)

\(= \lambda((-5) \times (-3) - 0 \times (-4)) - (-10)((-7) \times (-3) - 0 \times 1) + (-5)((-7) \times (-4) - 1 \times (-5)) = 0\)

\(= \lambda(15) + 10(21) - 5(33) = 0\)

\([\bar{a} \, \bar{b} \, \bar{c}] = 15\lambda + 45\) .........(2)

From (1) and (2):
\(15\lambda + 45 = 0\)
\(15\lambda = -45\)
\(\lambda = -3\)

(iii) For the given vectors:
\(\bar{a} = \hat{i} - \hat{j} + \hat{k}\)
\(\bar{b} = 2\hat{i} + \hat{j} - \hat{k}\)
\(\bar{c} = \lambda\hat{i} - \hat{j} + \lambda\hat{k}\)

Since the vectors are coplanar:
\([\bar{a} \, \bar{b} \, \bar{c}] = 0\) .........(1)

\(\begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{vmatrix} = 0\)

\(= 1(1 \times \lambda - (-1) \times (-1)) - (-1)(2 \times \lambda - (-1) \times \lambda) + 1(2 \times (-1) - \lambda \times 1) = 0\)

\(= 1(\lambda - 1) + 1(3\lambda) + 1(-\lambda - 2) = 0\)

\(= \lambda - 1 + 3\lambda - \lambda - 2 = 0\)

\([\bar{a} \, \bar{b} \, \bar{c}] = 3\lambda - 3\) .........(2)

From (1) and (2):
\(3\lambda - 3 = 0\)
\(3\lambda = 3\)
\(\lambda = 1\)
In simple words: Set the scalar triple product equal to zero and expand the determinant to find the value of λ. Simplify the resulting equation and solve for λ.

Exam Tip: When finding λ for coplanar vectors, always remember to set the scalar triple product equal to zero first - this is your starting condition. Expand the determinant step-by-step to avoid calculation errors.

Question 6. If \(\bar{a} = (2\hat{i} - \hat{j} + \hat{k})\), \(\bar{b} = (\hat{i} - 3\hat{j} - 5\hat{k})\) and \(\bar{c} = (3\hat{i} - 4\hat{j} - \hat{k})\), find \([\bar{a} \, \bar{b} \, \bar{c}]\) and interpret the result.
Answer: For the given vectors:
\(\bar{a} = 2\hat{i} - \hat{j} + \hat{k}\)
\(\bar{b} = \hat{i} - 3\hat{j} - 5\hat{k}\)
\(\bar{c} = 3\hat{i} - 4\hat{j} - \hat{k}\)

The scalar triple product is computed using the determinant:
\([\bar{a} \, \bar{b} \, \bar{c}] = \begin{vmatrix} 2 & -1 & 1 \\ 1 & -3 & -5 \\ 3 & -4 & -1 \end{vmatrix}\)

\(= 2((-3) \times (-1) - (-4) \times (-5)) - (-1)((-1) \times (-1) - 3 \times (-5)) + 1((-1) \times (-4) - 3 \times (-3))\)

\(= 2(-17) + 1(14) + 1(5)\)
\(= -34 + 14 + 5\)
\(= -15\)

The scalar triple product is \(-15\), which is not equal to zero. This means the vectors are not coplanar. The absolute value of the scalar triple product, which is 15 cubic units, represents the volume of the parallelepiped formed by the three vectors.
In simple words: The three vectors form a three-dimensional shape called a parallelepiped, and its volume is 15 cubic units. Since the product is not zero, the vectors do not lie in the same plane.

Exam Tip: Interpret the result of the scalar triple product clearly - if it is zero, the vectors are coplanar; if non-zero, its absolute value gives the volume of the parallelepiped formed by the three vectors.

Question 7. The volume of the parallelepiped whose edges are \((-12\hat{i} + \lambda\hat{k})\), \((3\hat{j} - \hat{k})\) and \((2\hat{i} + \hat{j} - 15\hat{k})\) is 546 cubic units. Find the value of λ.
Answer: Given:
1) Coterminous edges of the parallelepiped are:
\(\bar{a} = -12\hat{i} + \lambda\hat{k}\)
\(\bar{b} = 3\hat{j} - \hat{k}\)
\(\bar{c} = 2\hat{i} + \hat{j} - 15\hat{k}\)

2) Volume of the parallelepiped is 546 cubic units.

The volume of a parallelepiped is given by the absolute value of the scalar triple product:
\(\bar{V} = [\bar{a} \, \bar{b} \, \bar{c}] = \begin{vmatrix} -12 & 0 & \lambda \\ 0 & 3 & -1 \\ 2 & 1 & -15 \end{vmatrix}\)

\(= -12(3 \times (-15) - 1 \times (-1)) - 0 + \lambda(0 \times 1 - 3 \times 2)\)

\(= -12(-45 + 1) + \lambda(-6)\)

\(= -12(-44) - 6\lambda\)

\(= 528 - 6\lambda\)

Given that the volume is 546 cubic units:
\(|528 - 6\lambda| = 546\)

This gives two possibilities:
\(528 - 6\lambda = 546\) or \(528 - 6\lambda = -546\)

From the first equation:
\(-6\lambda = 18\)
\(\lambda = -3\)

From the second equation:
\(-6\lambda = -1074\)
\(\lambda = 179\)

Since the standard solution is \(\lambda = -3\).
In simple words: The volume formula uses the scalar triple product of the three edge vectors. You calculate the determinant, then set its absolute value equal to 546 and solve for λ.

Exam Tip: When finding unknown parameters in volume problems, remember that volume is always positive. Take the absolute value of the scalar triple product and solve the resulting equation - sometimes two solutions are possible, so check which makes sense in context.

Question 8. Show that the vectors \(\bar{a} = (\hat{i} + 3\hat{j} + \hat{k})\), \(\bar{b} = (2\hat{i} - \hat{j} - \hat{k})\) and \(\bar{c} = (7\hat{j} + 3\hat{k})\) are parallel to the same plane.
Answer: Given vectors:
\(\bar{a} = \hat{i} + 3\hat{j} + \hat{k}\)
\(\bar{b} = 2\hat{i} - \hat{j} - \hat{k}\)
\(\bar{c} = 7\hat{j} + 3\hat{k}\)

Vectors are parallel to the same plane if and only if they are coplanar. Three vectors are coplanar when their scalar triple product equals zero, so we must show that \([\bar{a} \, \bar{b} \, \bar{c}] = 0\).

\([\bar{a} \, \bar{b} \, \bar{c}] = \begin{vmatrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & 7 & 3 \end{vmatrix}\)

\(= 1((-1) \times 3 - 7 \times (-1)) - 3(2 \times 3 - 0 \times (-1)) + 1(2 \times 7 - 0 \times (-1))\)

\(= 1(-3 + 7) - 3(6) + 1(14)\)

\(= 4 - 18 + 14\)

\(= 0\)

Since the scalar triple product is zero, the vectors are coplanar. Therefore, the vectors lie in the same plane, which means they are all parallel to that plane.
In simple words: Three vectors are parallel to the same plane when they all fit together in one flat surface with no volume between them. This happens when the scalar triple product equals zero.

Exam Tip: The hint provided asks you to show that the scalar triple product is zero. Always expand this determinant methodically and verify that your final answer is indeed zero to confirm coplanarity.

Question 9. If the vectors \( \left( a\hat{i} + a\hat{j} + c\hat{k} \right) \), \( \left( \hat{i} + \hat{k} \right) \) and \( \left( c\hat{i} + c\hat{j} + b\hat{k} \right) \) be coplanar, show that \( c^2 = ab \).
Answer: Three vectors are coplanar when their scalar triple product equals zero. For the given vectors with components \( \bar{a} = a\hat{i} + a\hat{j} + c\hat{k} \), \( \bar{b} = \hat{i} + \hat{k} \), and \( \bar{c} = c\hat{i} + c\hat{j} + b\hat{k} \), we evaluate the determinant:
\[ \left[ \bar{a} \, \bar{b} \, \bar{c} \right] = \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0 \]
Expanding the determinant: \( a(0 \cdot b - c \cdot 1) - a(1 \cdot b - 1 \cdot c) + c(1 \cdot c - 0 \cdot c) = 0 \)
\( a(-c) - a(b - c) + c(c) = 0 \)
\( -ac - ab + ac + c^2 = 0 \)
\( c^2 - ab = 0 \)
\( c^2 = ab \)
In simple words: When three vectors lie in the same plane, the determinant made from their components must be zero. Working through the calculation for these specific vectors leads directly to the relation \( c^2 = ab \).

Exam Tip: Always set the scalar triple product equal to zero for coplanarity and expand the determinant carefully to avoid sign errors in the middle term.

Question 10. Show that the four points with position vectors \( \left( 4\hat{i} + 8\hat{j} + 12\hat{k} \right) \), \( \left( 2\hat{i} + 4\hat{j} + 6\hat{k} \right) \), \( \left( 3\hat{i} + 5\hat{j} + 4\hat{k} \right) \) and \( \left( 5\hat{i} + 8\hat{j} + 5\hat{k} \right) \) are coplanar.
Answer: For four points to be coplanar, the vectors formed by joining one point to the other three must satisfy the coplanarity condition. Let A, B, C, D denote the four points with the given position vectors. The vectors are:
\( \overline{BA} = 2\hat{i} + 4\hat{j} + 6\hat{k} \)
\( \overline{CA} = \hat{i} + 3\hat{j} + 8\hat{k} \)
\( \overline{DA} = -\hat{i} + 0\hat{j} + 7\hat{k} \)
The scalar triple product is:
\[ \left[ \overline{BA} \, \overline{CA} \, \overline{DA} \right] = \begin{vmatrix} 2 & 4 & 6 \\ 1 & 3 & 8 \\ -1 & 0 & 7 \end{vmatrix} = 0 \]
Expanding: \( 2(3 \times 7 - 0 \times 8) - 4(1 \times 7 - (-1) \times 8) + 6(1 \times 0 - (-1) \times 3) = 0 \)
\( 2(21) - 4(15) + 6(3) = 42 - 60 + 18 = 0 \)
Since the scalar triple product is zero, the vectors are coplanar, and therefore the four points lie in the same plane.
In simple words: Build vectors connecting one point to each of the other three, then check if their scalar triple product is zero. When it is, all four points sit on the same flat surface.

Exam Tip: Remember to compute vectors from one fixed point to the other three, and watch signs carefully when subtracting coordinates.

Question 11. Show that the four points with position vectors \( \left( 6\hat{i} - 7\hat{j} \right) \), \( \left( 16\hat{i} - 19\hat{j} - 4\hat{k} \right) \), \( \left( 3\hat{j} - 6\hat{k} \right) \) and \( \left( 2\hat{i} - 5\hat{j} + 10\hat{k} \right) \) are coplanar.
Answer: Let A, B, C, D be the four points corresponding to the given position vectors. We form vectors from point A to the other three points:
\( \overline{BA} = -10\hat{i} + 12\hat{j} + 4\hat{k} \)
\( \overline{CA} = 6\hat{i} - 10\hat{j} + 6\hat{k} \)
\( \overline{DA} = 4\hat{i} - 2\hat{j} - 10\hat{k} \)
We evaluate the scalar triple product:
\[ \left[ \overline{BA} \, \overline{CA} \, \overline{DA} \right] = \begin{vmatrix} -10 & 12 & 4 \\ 6 & -10 & 6 \\ 4 & -2 & -10 \end{vmatrix} = 0 \]
Expanding along the first row: \( -10((-10) \times (-10) - (-2) \times 6) - 12(6 \times (-10) - 4 \times 6) + 4(6 \times (-2) - 4 \times (-10)) = 0 \)
\( -10(112) - 12(-84) + 4(28) = -1120 + 1008 + 112 = 0 \)
Since the scalar triple product equals zero, the three vectors are coplanar, confirming that all four points lie in the same plane.
In simple words: Construct vectors from one point to each of the remaining three points, then compute their scalar triple product. When this product is zero, the four points are coplanar.

Exam Tip: Be careful with negative signs when forming vectors and when calculating determinants; a single sign error will change the final result.

Question 12. Find the value of \( \lambda \) for which the four points with position vectors \( \left( \hat{i} + 2\hat{j} + 3\hat{k} \right) \), \( \left( 3\hat{i} - \hat{j} + 2\hat{k} \right) \), \( \left( -2\hat{i} + \lambda\hat{j} + \hat{k} \right) \) and \( \left( 6\hat{i} - 4\hat{j} + 2\hat{k} \right) \) are coplanar.
Answer: Four points are coplanar when the scalar triple product of the three vectors formed from one point to the other three equals zero. Let A, B, C, D be the four points with the given position vectors. The vectors connecting from A to B, C, D are:
\( \overline{BA} = -2\hat{i} + 3\hat{j} + \hat{k} \)
\( \overline{CA} = 3\hat{i} + (2 - \lambda)\hat{j} + 2\hat{k} \)
\( \overline{DA} = -5\hat{i} + 6\hat{j} + \hat{k} \)
Setting up the determinant:
\[ \begin{vmatrix} -2 & 3 & 1 \\ 3 & (2 - \lambda) & 2 \\ -5 & 6 & 1 \end{vmatrix} = 0 \]
Expanding: \( -2((2 - \lambda) - 12) - 3(3 + 10) + 1(18 + 5(2 - \lambda)) = 0 \)
\( -2(2 - \lambda - 10) - 3(13) + 1(28 - 5\lambda) = 0 \)
\( 2\lambda + 20 - 39 + 28 - 5\lambda = 0 \)
\( 9 - 3\lambda = 0 \)
\( 3\lambda = 9 \)
\( \lambda = 3 \)
In simple words: To find the unknown parameter, set the determinant equal to zero and solve for \( \lambda \). This gives the unique value that makes all four points lie in a common plane.

Exam Tip: When a parameter appears in the determinant, expand carefully and collect all terms with that parameter on one side to solve the resulting equation.

Question 13. Find the value of \( \lambda \) for which the four points with position vectors \( \left( -\hat{j} + \hat{k} \right) \), \( \left( 2\hat{i} - \hat{j} - \hat{k} \right) \), \( \left( \hat{i} + \lambda\hat{j} + \hat{k} \right) \) and \( \left( 3\hat{j} + 3\hat{k} \right) \) are coplanar.
Answer: For four points to lie in the same plane, their scalar triple product must be zero. Let A, B, C, D denote the four points. We form vectors from A to the other three:
\( \overline{BA} = -2\hat{i} + 0\hat{j} + 2\hat{k} \)
\( \overline{CA} = -\hat{i} + (-1 - \lambda)\hat{j} + 0\hat{k} \)
\( \overline{DA} = 0\hat{i} - 4\hat{j} - 2\hat{k} \)
We set up the determinant and require it to equal zero:
\[ \begin{vmatrix} -2 & 0 & 2 \\ -1 & (-1 - \lambda) & 0 \\ 0 & -4 & -2 \end{vmatrix} = 0 \]
Expanding along the first row: \( -2((-1 - \lambda)(-2) - 0 \times (-4)) - 0 + 2((-1)(-4) - 0 \times (-1 - \lambda)) = 0 \)
\( -2(2 + 2\lambda) + 2(4) = 0 \)
\( -4 - 4\lambda + 8 = 0 \)
\( 4 - 4\lambda = 0 \)
\( \lambda = 1 \)
In simple words: Build the determinant using the components of the three vectors, set it to zero, and solve for \( \lambda \). The answer is the value that satisfies this equation.

Exam Tip: When expanding a determinant with a parameter, focus on rows or columns with many zeros to simplify calculations.

Question 14. Using vector method, show that the points A(4, 5, 1), B(0, -1, -1), C(3, 9, 4) and D(-4, 4, 4) are coplanar.
Answer: The position vectors of the given points are:
\( \bar{a} = 4\hat{i} + 5\hat{j} + \hat{k} \)
\( \bar{b} = 0\hat{i} - \hat{j} - \hat{k} \)
\( \bar{c} = 3\hat{i} + 9\hat{j} + 4\hat{k} \)
\( \bar{d} = -4\hat{i} + 4\hat{j} + 4\hat{k} \)
We form vectors from A to the other three points:
\( \overline{BA} = \bar{a} - \bar{b} = 4\hat{i} + 6\hat{j} + 2\hat{k} \)
\( \overline{CA} = \bar{a} - \bar{c} = \hat{i} - 4\hat{j} - 3\hat{k} \)
\( \overline{DA} = \bar{a} - \bar{d} = 8\hat{i} + \hat{j} - 3\hat{k} \)
To verify coplanarity, we compute the scalar triple product:
\[ \left[ \overline{BA} \, \overline{CA} \, \overline{DA} \right] = \begin{vmatrix} 4 & 6 & 2 \\ 1 & -4 & -3 \\ 8 & 1 & -3 \end{vmatrix} \]
Expanding along the first row: \( 4((-4)(-3) - 1 \times (-3)) - 6(1 \times (-3) - 8 \times (-3)) + 2(1 \times 1 - 8 \times (-4)) = 0 \)
\( 4(12 + 3) - 6(-3 + 24) + 2(1 + 32) = 4(15) - 6(21) + 2(33) = 60 - 126 + 66 = 0 \)
Since the scalar triple product is zero, the four points lie in a common plane.
In simple words: Extract position vectors from the given coordinates, form three vectors from one point to the others, and check if their scalar triple product is zero. If it is, all four points are coplanar.

Exam Tip: When given points in coordinate form, always write their position vectors first, then construct difference vectors systematically to ensure no arithmetic errors.

Question 15. Find the value of λ for which the points A(3, 2, 1), B(4, λ, 5), C(4, 2, -2) and D(6, 5, -1) are coplanar.
Answer: Given points A ≡ (3, 2, 1), B ≡ (4, λ, 5), C ≡ (4, 2, -2), and D ≡ (6, 5, -1). The position vectors are \( \overline{a} = 3i + 2j + k \), \( \overline{b} = 4i + λj + 5k \), \( \overline{c} = 4i + 2j - 2k \), and \( \overline{d} = 6i + 5j - k \).

The vectors from A are: \( \overline{BA} = -i + (2-λ)j - 4k \), \( \overline{CA} = -i + 0j + 3k \), and \( \overline{DA} = -3i - 3j + 2k \).

For coplanarity, the scalar triple product must equal zero:
\[ \begin{vmatrix} -1 & (2-λ) & -4 \\ -1 & 0 & 3 \\ -3 & -3 & 2 \end{vmatrix} = 0 \]

Expanding: \( -1(0 \times 2 - 3 × (-3)) - (2-λ)((-1) × 2 - (-3) × (-3)) - 4((-1) × (-3) - (-3) × 0) = 0 \)

This simplifies to: \( -1(9) - (2-λ)(7) - 4(3) = 0 \)

\( -9 - 14 + 7λ - 12 = 0 \)

\( 7λ - 35 = 0 \)

\( λ = 5 \)

In simple words: Four points lie on the same plane when their position vectors create a special arrangement - the scalar triple product becomes zero. By setting up this condition with the given coordinates and solving, we find that λ must be 5.

Exam Tip: Always use the scalar triple product formula for coplanarity - remember to compute the determinant carefully by expanding along one row, and double-check your arithmetic when substituting values.

Exercise 25B

Question 1. If \( \overline{a} = xi + 2j - zk \) and \( \overline{b} = 3i - yj + k \) are two equal vectors, then x + y + z = ?
Answer: Since the two vectors are equal, their corresponding components must match. Comparing each component: the i-component gives x = 3; the j-component gives 2 = -y, so y = -2; the k-component gives -z = 1, so z = -1.

Therefore, \( x + y + z = 3 + (-2) + (-1) = 0 \)

In simple words: When two vectors are equal, every part of one must match the same part of the other. Once you match the parts, just add the three numbers together.

Exam Tip: Equal vectors have identical components in every direction - compare i, j, and k separately before combining your answer.

Question 2. Write a unit vector in the direction of the sum of the vectors \( \overline{a} = (2i + 2j - 5k) \) and \( \overline{b} = (2i + j - 7k) \).
Answer: The sum of the vectors is \( \overline{s} = \overline{a} + \overline{b} = 2i + 2j - 5k + 2i + j - 7k = 4i + 3j - 12k \).

The magnitude of this sum is \( |\overline{s}| = \sqrt{4^2 + 3^2 + (-12)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \).

The unit vector is found by dividing each component by the magnitude: \( \hat{s} = \frac{\overline{s}}{|\overline{s}|} = \frac{4i + 3j - 12k}{13} \)

In simple words: Add the two vectors together to get one new vector. Then find how long that vector is. Finally, divide each part of the vector by its length - this creates a unit vector pointing in the same direction.

Exam Tip: A unit vector always has magnitude 1 - verify this by checking that \( |\hat{s}| = 1 \) after computing it.

Question 3. Write the value of λ so that the vectors \( \overline{a} = (2i + λj + k) \) and \( \overline{b} = (i - 2j + 3k) \) are perpendicular to each other.
Answer: Two vectors are perpendicular when their dot product equals zero. The dot product is: \( \overline{a} \cdot \overline{b} = (2)(1) + (λ)(-2) + (1)(3) = 2 - 2λ + 3 = 5 - 2λ \).

Setting this equal to zero for perpendicularity: \( 5 - 2λ = 0 \)

\( λ = \frac{5}{2} \)

In simple words: Perpendicular vectors are at right angles, and this happens when their dot product is zero. By computing the dot product and setting it to zero, we can solve for the unknown number λ.

Exam Tip: Always verify perpendicularity by confirming the dot product is exactly zero - this is the defining property.

Question 4. Find the value of p for which the vectors \( \overline{a} = (3i + 2j + 9k) \) and \( \overline{b} = (i - 2pj + 3k) \) are parallel.
Answer: Two vectors are parallel when one is a scalar multiple of the other - that is, their components are proportional. This gives us: \( \frac{3}{1} = \frac{2}{-2p} = \frac{9}{3} \).

From the first and third ratios: \( 3 = 3 \) ✓

From the first two ratios: \( 3 = \frac{2}{-2p} \)

\( 3 × (-2p) = 2 \)

\( -6p = 2 \)

\( p = -\frac{1}{3} \)

In simple words: Parallel vectors point in the same (or opposite) direction. This means each component of one vector must be the same multiple of the corresponding component of the other vector.

Exam Tip: Set up the proportion of all three component ratios and solve - they must all give the same value for the vectors to be parallel.

Question 5. Find the value of λ when the projection of \( \overline{a} = (λi + j + 4k) \) on \( \overline{b} = (2i + 6j + 3k) \) is 4 units.
Answer: The projection of \( \overline{a} \) onto \( \overline{b} \) is given by \( \text{proj} = \frac{\overline{a} \cdot \overline{b}}{|\overline{b}|} \).

First, calculate \( |\overline{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \).

The dot product is \( \overline{a} \cdot \overline{b} = λ(2) + (1)(6) + (4)(3) = 2λ + 6 + 12 = 2λ + 18 \).

Setting the projection equal to 4: \( \frac{2λ + 18}{7} = 4 \)

\( 2λ + 18 = 28 \)

\( 2λ = 10 \)

\( λ = 5 \)

In simple words: The projection is like the shadow one vector casts on another. To find it, compute the dot product and divide by the length of the vector being projected onto.

Exam Tip: Always compute the magnitude of the base vector first - this prevents careless errors in the division step.

Question 6. If \( \overline{a} \) and \( \overline{b} \) are perpendicular vectors such that \( |\overline{a} + \overline{b}| = 13 \) and \( |\overline{a}| = 5 \), find the value of \( |\overline{b}| \).
Answer: Since \( \overline{a} \) and \( \overline{b} \) are perpendicular, their dot product is zero: \( \overline{a} \cdot \overline{b} = 0 \).

Using the magnitude formula for the sum: \( |\overline{a} + \overline{b}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + 2\overline{a} \cdot \overline{b} \).

Substituting the known values: \( 13^2 = 5^2 + |\overline{b}|^2 + 2(0) \)

\( 169 = 25 + |\overline{b}|^2 \)

\( |\overline{b}|^2 = 144 \)

\( |\overline{b}| = 12 \)

In simple words: When two perpendicular vectors are added, the Pythagorean theorem applies - the magnitude of the sum relates to the magnitudes of the parts through a familiar square relationship.

Exam Tip: For perpendicular vectors, always use the simplified magnitude formula since the cross term vanishes - this makes calculation much cleaner.

Question 7. If \( \overline{a} \) is a unit vector such that \( (\overline{x} - \overline{a}) \cdot (\overline{x} + \overline{a}) = 15 \), find \( |\overline{x}| \).
Answer: Expanding the dot product using the difference of squares pattern: \( (\overline{x} - \overline{a}) \cdot (\overline{x} + \overline{a}) = |\overline{x}|^2 - |\overline{a}|^2 \).

Since \( \overline{a} \) is a unit vector, \( |\overline{a}| = 1 \), so \( |\overline{a}|^2 = 1 \).

Therefore: \( |\overline{x}|^2 - 1 = 15 \)

\( |\overline{x}|^2 = 16 \)

\( |\overline{x}| = 4 \)

In simple words: The expression expands like (p - q)(p + q) = p² - q². Since we know one piece (the unit vector's magnitude is 1), we can easily solve for the other.

Exam Tip: Recognize algebraic patterns in vector products - the difference of squares simplifies calculations dramatically.

Question 8. Find the sum of the vectors \( \overline{a} = (i - 3k) \), \( \overline{b} = (2j - k) \), and \( \overline{c} = (2i - 3j + 2k) \).
Answer: Adding the vectors component by component:
\( \overline{a} + \overline{b} + \overline{c} = (1 + 0 + 2)i + (0 + 2 - 3)j + (-3 - 1 + 2)k = 3i - j - 2k \)

In simple words: Combine all the i-parts, all the j-parts, and all the k-parts separately to get the total vector.

Exam Tip: Line up vectors by component and add vertically - organizing your work this way prevents sign errors.

Question 9. Find the sum of the vectors \( \overline{a} = (i - 2j) \), \( \overline{b} = (2i - 3j) \), and \( \overline{c} = (2i + 3k) \).
Answer: Adding component by component: \( \overline{a} + \overline{b} + \overline{c} = (1 + 2 + 2)i + (-2 - 3 + 0)j + (0 + 0 + 3)k = 5i - 5j + 3k \)

In simple words: Gather all the horizontal parts, all the forward parts, and all the vertical parts to get one combined vector.

Exam Tip: Check that you've accounted for all three components - if a component doesn't appear in a vector, treat it as zero.

Question 10. Write the projection of the vector \( (i + j + k) \) along the vector \( j \).
Answer: The projection of \( \overline{a} = i + j + k \) onto \( \overline{b} = j \) is calculated as: \( \text{proj} = \frac{\overline{a} \cdot \overline{b}}{|\overline{b}|} = \frac{(1)(0) + (1)(1) + (1)(0)}{1} = 1 \)

In simple words: The projection tells you how much of the first vector points in the direction of the second vector - here only the j-component of the first vector contributes.

Exam Tip: Unit vectors in the coordinate directions simplify the projection calculation - the result is just the corresponding component.

Question 11. Write the projection of the vector \( (7i + j - 4k) \) on the vector \( (2i + 6j + 3k) \).
Answer: Calculate the magnitude: \( |\overline{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = 7 \).

The dot product is: \( \overline{a} \cdot \overline{b} = (7)(2) + (1)(6) + (-4)(3) = 14 + 6 - 12 = 8 \).

The projection is: \( \text{proj} = \frac{8}{7} \)

In simple words: Multiply matching parts of the two vectors and add them up. Then divide by how long the second vector is.

Exam Tip: Double-check the dot product arithmetic - a single sign error ruins the entire calculation.

Question 12. Find \( \overline{a} \cdot (\overline{b} \times \overline{c}) \) when \( \overline{a} = (2i + j + 3k) \), \( \overline{b} = (-i + 2j + k) \), and \( \overline{c} = (3i + j + 2k) \).
Answer: First, calculate the cross product \( \overline{b} \times \overline{c} \):
\[ \overline{b} \times \overline{c} = \begin{vmatrix} i & j & k \\ -1 & 2 & 1 \\ 3 & 1 & 2 \end{vmatrix} = i(4-1) - j(-2-3) + k(-1-6) = 3i + 5j - 7k \]

Now compute the dot product: \( \overline{a} \cdot (\overline{b} \times \overline{c}) = (2)(3) + (1)(5) + (3)(-7) = 6 + 5 - 21 = -10 \)

In simple words: First find the cross product (which creates a perpendicular vector), then multiply that result by dot product with the first vector. The final answer tells you the volume-related property of the three vectors.

Exam Tip: The scalar triple product is order-dependent - always compute the cross product first, then the dot product with the remaining vector.

Question 13. Find a vector in the direction of \( (2i - 3j + 6k) \) which has magnitude 21 units.
Answer: Calculate the magnitude of the given vector: \( |\overline{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = 7 \).

The unit vector is: \( \hat{a} = \frac{2i - 3j + 6k}{7} \).

A vector of magnitude 21 in the same direction is: \( 21 \times \frac{2i - 3j + 6k}{7} = 3(2i - 3j + 6k) = 6i - 9j + 18k \)

In simple words: Find how long the original vector is. Then create a unit vector by dividing by that length. Finally, stretch it to the desired length by multiplying.

Exam Tip: Verify your answer by checking that its magnitude is exactly 21 - this catches calculation errors immediately.

Question 14. If \( \overline{a} = (2i + 2j + 3k) \), \( \overline{b} = (-i + 2j + k) \), and \( \overline{c} = (3i + j) \) are such that \( (\overline{a} + λ\overline{b}) \) is perpendicular to \( \overline{c} \), then find the value of λ.
Answer: Form the expression: \( \overline{a} + λ\overline{b} = 2i + 2j + 3k + λ(-i + 2j + k) = (2-λ)i + (2+2λ)j + (3+λ)k \).

For perpendicularity to \( \overline{c} \), the dot product must be zero: \( ((2-λ)i + (2+2λ)j + (3+λ)k) \cdot (3i + j) = 0 \)

\( 3(2-λ) + 1(2+2λ) = 0 \)

\( 6 - 3λ + 2 + 2λ = 0 \)

\( 8 - λ = 0 \)

\( λ = 8 \)

In simple words: Construct the combined vector using the unknown λ. Then use the perpendicularity condition (dot product equals zero) to solve for λ.

Exam Tip: Always expand the vector expression completely before computing the dot product - partial expansion often leads to missed terms.

Question 15. Write the vector of magnitude 15 units in the direction of vector \( (i - 2j + 2k) \).
Answer: Find the magnitude: \( |\overline{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3 \).

The unit vector is: \( \hat{a} = \frac{i - 2j + 2k}{3} \).

Multiply by 15 to get the required vector: \( 15 \times \frac{i - 2j + 2k}{3} = 5(i - 2j + 2k) = 5i - 10j + 10k \)

In simple words: Scale the original vector so it points the same way but has exactly 15 units of length - do this by finding the unit vector first, then stretching it.

Exam Tip: Check that \( |15 \text{ vector}| = 15 \) by computing \( \sqrt{5^2 + (-10)^2 + 10^2} \) to verify correctness.

Question 16. If \( \overline{a} = (i + j + k) \), \( \overline{b} = (4i - 2j + 3k) \), and \( \overline{c} = (i - 2j + k) \), find a vector of magnitude 6 units which is parallel to the vector \( (2\overline{a} - \overline{b} + 3\overline{c}) \).
Answer: Calculate the expression: \( 2\overline{a} - \overline{b} + 3\overline{c} = 2(i + j + k) - (4i - 2j + 3k) + 3(i - 2j + k) = 2i + 2j + 2k - 4i + 2j - 3k + 3i - 6j + 3k = i - 2j + 2k \).

Find its magnitude: \( |i - 2j + 2k| = \sqrt{1 + 4 + 4} = 3 \).

The unit vector is: \( \frac{i - 2j + 2k}{3} \).

Scale to magnitude 6: \( 6 \times \frac{i - 2j + 2k}{3} = 2(i - 2j + 2k) = 2i - 4j + 4k \)

In simple words: Assemble the combined vector by following the algebraic expression carefully. Then resize it to the target magnitude using the unit vector method.

Exam Tip: Combine like terms systematically when building the final vector expression - organize by i, j, k components to avoid losing terms.

Question 17. Write the projection of the vector \( (i - j) \) on the vector \( (i + j) \).
Answer: Calculate the magnitude: \( |\overline{b}| = \sqrt{1^2 + 1^2} = \sqrt{2} \).

The unit vector is: \( \hat{b} = \frac{i + j}{\sqrt{2}} \).

Compute the dot product: \( \overline{a} \cdot \overline{b} = (1)(1) + (-1)(1) = 1 - 1 = 0 \).

Therefore, the projection is: \( \frac{0}{\sqrt{2}} = 0 \)

In simple words: When the dot product is zero, the two vectors are perpendicular, meaning one casts no shadow onto the other.

Exam Tip: A zero projection indicates perpendicularity - always verify this makes geometric sense given the vectors.

Question 18. Write the angle between two vectors \( \overline{a} \) and \( \overline{b} \) with magnitudes \( \sqrt{3} \) and 2 respectively, having \( \overline{a} \cdot \overline{b} = \sqrt{6} \).
Answer: Use the dot product formula: \( \overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}|\cosθ \).

Substituting the values: \( \sqrt{6} = \sqrt{3} \times 2 \times \cosθ \)

\( \sqrt{6} = 2\sqrt{3}\cosθ \)

\( \cosθ = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)

\( θ = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45° = \frac{π}{4} \)

In simple words: The angle between vectors relates to their dot product through a cosine formula. Once you know the magnitudes and dot product, you can find the angle using inverse cosine.

Exam Tip: Simplify the fraction under the cosine - recognizing standard values like \( \frac{1}{\sqrt{2}} \) saves time and reduces errors.

Question 19. If \( \overline{a} = (i - 7j + 7k) \) and \( \overline{b} = (3i - 2j + 2k) \), then find \( |\overline{a} \times \overline{b}| \).
Answer: Compute the cross product:
\[ \overline{a} \times \overline{b} = \begin{vmatrix} i & j & k \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix} = i((-7)(2) - (7)(-2)) - j((1)(2) - (7)(3)) + k((1)(-2) - (-7)(3)) \]

\( = i(-14 + 14) - j(2 - 21) + k(-2 + 21) = 0i + 19j + 19k \)

The magnitude is: \( |\overline{a} \times \overline{b}| = \sqrt{0^2 + 19^2 + 19^2} = \sqrt{2 \times 19^2} = 19\sqrt{2} \)

In simple words: The cross product creates a new vector perpendicular to both original vectors. Its magnitude represents the area of the parallelogram formed by the two vectors.

Exam Tip: Factor out perfect squares when simplifying - \( \sqrt{2 \times 19^2} = 19\sqrt{2} \) is much cleaner than leaving the radical unsimplified.

Question 20. Find the angle between two vectors \( \overline{a} \) and \( \overline{b} \) with magnitudes 1 and 2 respectively, when \( |\overline{a} \times \overline{b}| = \sqrt{3} \).
Answer: Use the cross product magnitude formula: \( |\overline{a} \times \overline{b}| = |\overline{a}||\overline{b}|\sinθ \).

Substituting: \( \sqrt{3} = 1 \times 2 \times \sinθ \)

\( \sinθ = \frac{\sqrt{3}}{2} \)

\( θ = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60° = \frac{π}{3} \)

In simple words: The magnitude of a cross product gives you information about the angle between vectors through a sine relationship. Larger cross product magnitudes mean the vectors are more perpendicular.

Exam Tip: Memorize the standard sine and cosine values (\( \sin 60° = \frac{\sqrt{3}}{2} \), \( \cos 45° = \frac{1}{\sqrt{2}} \), etc.) to avoid calculator errors on timed exams.

Question 21. What conclusion can you draw about vectors \( \vec{a} \) and \( \vec{b} \) when \( \vec{a} \times \vec{b} = \vec{0} \) and \( \vec{a} \cdot \vec{b} = \vec{0} \)?
Answer: Given that \( \vec{a} \times \vec{b} = \vec{0} \) and \( \vec{a} \cdot \vec{b} = \vec{0} \), we have:
\( |\vec{a}||\vec{b}|\sin\theta = |\vec{a}||\vec{b}|\cos\theta = \vec{0} \)
Since sine and cosine cannot simultaneously equal zero, \( |\vec{a}| = |\vec{b}| = \vec{0} \)
Conclusion: When \( \vec{a} \times \vec{b} = \vec{0} \) and \( \vec{a} \cdot \vec{b} = \vec{0} \), then \( |\vec{a}| = |\vec{b}| = 0 \). This means both vectors are null vectors.
In simple words: If the cross product and dot product of two vectors are both zero, then at least one of them must be a zero vector.

Exam Tip: Remember that the cross product being zero indicates parallel vectors and the dot product being zero indicates perpendicular vectors - these conditions can only both hold if at least one vector is zero.

Question 22. Find the value of λ when the vectors \( \vec{a} = \left(\hat{i} + \lambda\hat{j} + 3\hat{k}\right) \) and \( \vec{b} = \left(3\hat{i} + 2\hat{j} + 9\hat{k}\right) \) are parallel.
Answer: Given:
\( \vec{a} = \hat{i} + \lambda\hat{j} + 3\hat{k} \)
\( \vec{b} = 3\hat{i} + 2\hat{j} + 9\hat{k} \)
For vectors to be parallel: \( \vec{a} \parallel \vec{b} \)
\( \frac{1}{3} = \frac{\lambda}{2} = \frac{3}{9} \)
\( \frac{1}{3} = \frac{\lambda}{2} \)
\( \lambda = 2 \times \frac{1}{3} = \frac{2}{3} \)
Therefore, \( \lambda = \frac{2}{3} \)
In simple words: Two vectors are parallel when their components are proportional. Setting up the ratio of corresponding components and solving gives us the value of λ.

Exam Tip: For parallel vectors, always use the condition that ratios of corresponding components must be equal.

Question 23. Write the value of \( \hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}) \).
Answer: We recognize the following results:
\( \hat{i} \times \hat{j} = \hat{k}, \hat{j} \times \hat{k} = \hat{i}, \hat{k} \times \hat{i} = \hat{j} \)
\( \hat{j} \times \hat{i} = -\hat{k}, \hat{k} \times \hat{j} = -\hat{i}, \hat{i} \times \hat{k} = -\hat{j} \)
\( \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \)
Therefore:
\( \hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}) = \hat{i} \cdot \hat{i} + \hat{j} \cdot (-\hat{j}) + \hat{k} \cdot \hat{k} = 1 - 1 + 1 = 1 \)
In simple words: Use the standard cross-product rules for unit vectors and their dot products to evaluate the expression step by step.

Exam Tip: Always memorize the basic cross-product results for standard unit vectors \( \hat{i}, \hat{j}, \hat{k} \) to evaluate such expressions efficiently.

Question 24. Find the volume of the parallelepiped whose edges are represented by the vectors \( \vec{a} = (2\hat{i} - 3\hat{j} + 4\hat{k}), \vec{b} = (\hat{i} + 2\hat{j} - \hat{k}) \) and \( \vec{c} = (3\hat{i} - 2\hat{j} + 2\hat{k}) \).
Answer: The scalar triple product geometrically gives the volume of the parallelepiped with three coterminous edges. The volume is \( V = |\vec{a} \cdot (\vec{b} \times \vec{c})| \)
\( \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \)
\( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \)
\( \vec{c} = 3\hat{i} - 2\hat{j} + 2\hat{k} \)
\( V = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -2 & 2 \end{vmatrix} = 2(4 - 2) - (-3)(2 - (-3)) + 4(-2 - 6) = 4 + 15 - 32 = |-13| = 13 \) cubic units
In simple words: To find the volume of a parallelepiped, set up the determinant of the three edge vectors and compute its absolute value.

Exam Tip: Always take the absolute value of the determinant since volume is always positive; a negative result indicates you need the magnitude.

Question 25. If \( \vec{a} = (-2\hat{i} - 2\hat{j} + 4\hat{k}), \vec{b} = (-2\hat{j} + 4\hat{j} - 2\hat{k}) \) and \( \vec{c} = (4\hat{i} - 2\hat{j} - 2\hat{k}) \), then prove that \( \vec{a}, \vec{b} \) and \( \vec{c} \) are coplanar.
Answer: Vectors are coplanar when their scalar triple product equals zero: \( [\vec{a} \vec{b} \vec{c}] = 0 \)
\( \vec{a} = -2\hat{i} - 2\hat{j} + 4\hat{k} \)
\( \vec{b} = -2\hat{i} + 4\hat{j} - 2\hat{k} \)
\( \vec{c} = 4\hat{i} - 2\hat{j} - 2\hat{k} \)
L.H.S. \( = \begin{vmatrix} -2 & -2 & 4 \\ -2 & 4 & -2 \\ 4 & -2 & -2 \end{vmatrix} = -2(-8 - 4) + 2(4 + 8) + 4(4 - 16) = 24 + 24 - 48 = 0 \) = R.H.S.
Since L.H.S. = R.H.S., the vectors \( \vec{a} = -2\hat{i} - 2\hat{j} + 4\hat{k} \), \( \vec{b} = -2\hat{i} + 4\hat{j} - 2\hat{k} \) and \( \vec{c} = 4\hat{i} - 2\hat{j} - 2\hat{k} \) are coplanar. Hence proved.
In simple words: Three vectors lie in the same plane (are coplanar) when their scalar triple product equals zero.

Exam Tip: To prove coplanarity, compute the scalar triple product as a determinant - if it equals zero, the vectors definitely lie in the same plane.

Question 26. If \( \vec{a} = (2\hat{i} + 6\hat{j} + 27\hat{k}) \) and \( \vec{b} = (\hat{i} + \lambda\hat{j} + \mu\hat{k}) \) are such that \( \vec{a} \times \vec{b} = \vec{0} \), then find the values of λ and μ.
Answer: Given:
\( \vec{a} = 2\hat{i} + 6\hat{j} + 27\hat{k} \)
\( \vec{b} = \hat{i} + \lambda\hat{j} + \mu\hat{k} \)
\( \vec{a} \times \vec{b} = \vec{0} \)
\( (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0} \)
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = 0 = \hat{i}(6\mu - 27\lambda) - \hat{j}(2\mu - 27) + \hat{k}(2\lambda - 6) \)
From the \( \hat{k} \) component: \( 2\lambda - 6 = 0 \) which gives \( \lambda = 3 \)
From the \( \hat{j} \) component: \( 2\mu - 27 = 0 \) which gives \( \mu = \frac{27}{2} \)
Therefore, \( \lambda = 3, \mu = \frac{27}{2} \)
In simple words: When the cross product of two vectors is zero, they are parallel, so the ratios of their components must be equal.

Exam Tip: For a cross product to be zero, all three components of the resulting vector must vanish - set each component equation equal to zero and solve.

Question 27. If θ is the angle between \( \vec{a} \) and \( \vec{b} \), and \( |\vec{a} \times \vec{b}| = |\vec{a} \cdot \vec{b}| \), then what is the value of θ?
Answer: Given that \( |\vec{a} \times \vec{b}| = |\vec{a} \cdot \vec{b}| \):
\( |\vec{a}||\vec{b}|\sin\theta = |\vec{a}||\vec{b}|\cos\theta \)
\( \sin\theta = \cos\theta \)
\( \tan\theta = 1 \)
\( \theta = \tan^{-1}(1) = \frac{\pi}{4} \)
Therefore, \( \theta = \frac{\pi}{4} \)
In simple words: When the magnitude of the cross product equals the magnitude of the dot product, the angle between the vectors is 45 degrees.

Exam Tip: Remember that \( \sin\theta = \cos\theta \) only when \( \theta = \frac{\pi}{4} \) or 45 degrees in the range \( [0, \pi] \).

Question 28. When does \( |\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| \) hold?
Answer: When two vectors are parallel or collinear, they combine in a scalar manner since the angle between them is zero degrees - they point in the same or opposite directions. Therefore, when two vectors \( \vec{a} \) and \( \vec{b} \) are either parallel or collinear, then \( |\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| \).
In simple words: This equality holds when both vectors point in exactly the same direction (they are parallel).

Exam Tip: Recall that vector addition follows the parallelogram law - only in the special case of parallel vectors does the magnitude of the sum equal the sum of magnitudes.

Question 29. Find the direction cosines of a vector which is equally inclined to the x - axis, y - axis and z - axis.
Answer: Let the inclination angles with the three coordinate axes be:
x - axis = \( \alpha \)
y - axis = \( \beta \)
z - axis = \( \gamma \)
Since the vector is equally inclined to all three axes: \( \alpha = \beta = \gamma \)
Direction cosines are: \( \cos\alpha, \cos\beta, \cos\gamma \)
Using the standard property: \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \)
Substituting \( \alpha = \beta = \gamma \):
\( \cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \)
\( 3\cos^2\alpha = 1 \)
\( \cos\alpha = \frac{1}{\sqrt{3}} \)
Therefore, \( \cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt{3}} \)
The direction cosines are \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)
In simple words: For a vector equally inclined to all three coordinate axes, each direction cosine has the same value, found using the constraint that their squares sum to 1.

Exam Tip: Always use the fundamental identity \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \) along with any given conditions about equal inclination.

Question 30. If P(1, 5, 4) and Q(4, 1, - 2) be the position vectors of two points P and Q, find the direction ratios of \( \overrightarrow{PQ} \).
Answer: Let P(\( x_1, y_1, z_1 \)) and Q(\( x_2, y_2, z_2 \)) represent two points. The direction ratios of the line connecting P and Q, denoted \( \overrightarrow{PQ} \), are \( x_2 - x_1, y_2 - y_1, z_2 - z_1 \)
Given: P is (1, 5, 4) and Q is (4, 1, - 2)
Direction ratios of \( \overrightarrow{PQ} \) are: (4 - 1), (1 - 5), (- 2 - 4) = 3, - 4, - 6
Therefore, the direction ratios of \( \overrightarrow{PQ} \) are: 3, - 4, - 6
In simple words: To find direction ratios between two points, subtract the coordinates of the first point from the corresponding coordinates of the second point.

Exam Tip: Direction ratios can be any scalar multiple of the actual differences in coordinates - simplifying by a common factor is acceptable.

Question 31. Find the direction cosines of the vector \( \vec{a} = (\hat{i} + 2\hat{j} + 3\hat{k}) \).
Answer: Given \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \)
Let the inclination angles with the three coordinate axes be:
x - axis = \( \alpha \)
y - axis = \( \beta \)
z - axis = \( \gamma \)
Direction cosines: \( \cos\alpha, \cos\beta, \cos\gamma = l, m, n \)
For a vector \( \vec{a} = a\hat{i} + b\hat{j} + c\hat{k} \):
\( l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} \)
\( l = \frac{1}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{1}{\sqrt{1 + 4 + 9}} = \frac{1}{\sqrt{14}} \)
\( m = \frac{2}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{2}{\sqrt{1 + 4 + 9}} = \frac{2}{\sqrt{14}} \)
\( n = \frac{3}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{1 + 4 + 9}} = \frac{3}{\sqrt{14}} \)
Therefore, the direction cosines are \( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \)
In simple words: Divide each component of the vector by its magnitude to obtain the direction cosines.

Exam Tip: Always compute the magnitude first, then divide each component by this magnitude to find direction cosines.

Question 32. If \( \hat{a} \) and \( \hat{b} \) are unit vectors such that \( (\hat{a} + \hat{b}) \) is a unit vector, what is the angle between \( \hat{a} \) and \( \hat{b} \)?
Answer: Given that \( \hat{a} \) and \( \hat{b} \) are unit vectors, and \( (\hat{a} + \hat{b}) \) is also a unit vector:
\( |\hat{a}| = |\hat{b}| = |\hat{a} + \hat{b}| = 1 \)
Since the modulus of a unit vector equals unity.
Now, \( |\hat{a} + \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2|\hat{a}||\hat{b}|\cos\theta \)
\( 1^2 = 1^2 + 1^2 + 2 \times 1 \times 1 \times \cos\theta \)
\( 1 = 1 + 1 + 2\cos\theta \)
\( \cos\theta = \frac{1 - 1 - 1}{2} = -\frac{1}{2} \)
\( \theta = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \)
Therefore, the angle is \( \frac{2\pi}{3} \)
In simple words: When the sum of two unit vectors is also a unit vector, the angle between them is 120 degrees.

Exam Tip: Use the magnitude formula for the sum of two vectors to set up an equation, then solve for the angle using the cosine formula.

Question 1 (Objective). Mark (√) against the correct answer: A unit vector in the direction of the vector \( \vec{a} = (2\hat{i} - 3\hat{j} + 6\hat{k}) \) is
(a) \( \left(\hat{i} - \frac{3}{2}\hat{j} + 3\hat{k}\right) \)
(b) \( \left(\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}\right) \)
(c) \( \left(\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}\right) \)
(d) none of these
Answer: (b) \( \left(\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}\right) \)
In simple words: To find a unit vector in the direction of any given vector, divide each component by the vector's magnitude. The magnitude of \( 2\hat{i} - 3\hat{j} + 6\hat{k} \) is \( \sqrt{4 + 9 + 36} = 7 \), so the unit vector is obtained by dividing each component by 7.

Exam Tip: Always calculate the magnitude using \( \sqrt{a^2 + b^2 + c^2} \) before dividing to get the unit vector.

Question 2 (Objective). Mark (√) against the correct answer: The direction cosines of the vector \( \vec{a} = (-2\hat{i} + \hat{j} - 5\hat{k}) \) are
(a) -2, 1, -5
(b) \( \frac{1}{3}, \frac{-1}{6}, \frac{-5}{6} \)
(c) \( \frac{2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}} \)
(d) \( \frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}} \)
Answer: (d) \( \frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}} \)
In simple words: The direction cosines of a vector \( a\hat{i} + b\hat{j} + c\hat{k} \) are \( \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}} \). For the given vector, the magnitude is \( \sqrt{4 + 1 + 25} = \sqrt{30} \).

Exam Tip: Verify your answer by checking that the sum of squares of direction cosines equals 1.

Question 3 (Objective). Mark (√) against the correct answer: If A(1, 2, -3) and B(-1, -2, 1) are the end points of a vector \( \overrightarrow{AB} \), then the direction cosines of \( \overrightarrow{AB} \) are
(a) -2, -4, 4
(b) \( \frac{-1}{3}, \frac{-2}{3}, \frac{2}{3} \)
(c) \( \frac{-1}{3}, \frac{-2}{3}, \frac{2}{3} \)
(d) none of these
Answer: (b) \( \frac{-1}{3}, \frac{-2}{3}, \frac{2}{3} \)
In simple words: The vector from A to B is found by subtracting A's coordinates from B's. This gives \( -2\hat{i} - 4\hat{j} + 4\hat{k} \), which has magnitude 6. Dividing by 6 gives the direction cosines.

Exam Tip: Always ensure you subtract in the correct order (endpoint minus starting point) when finding the direction of a vector between two points.

Question 4 (Objective). Mark (√) against the correct answer: If a vector makes angle α, β and γ with the x-axis, y-axis and z-axis respectively, then the value of (sin²α + sin²β + sin²γ) is
(a) 1
(b) 2
(c) 0
(d) 3
Answer: (b) 2
In simple words: Starting with the identity \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \), substitute \( \sin^2\theta = 1 - \cos^2\theta \) for each angle to get the sum of the sine squares.

Exam Tip: Use the fundamental relationship between sine and cosine squared - this problem relies on converting the cosine identity into a sine identity.

Question 5 (Objective). Mark (√) against the correct answer: The vector \( (\cos\alpha\cos\beta)\hat{i} + (\cos\alpha\sin\beta)\hat{j} + (\sin\alpha)\hat{k} \) is a
(a) null vector
(b) unit vector
(c) a constant vector
(d) none of these
Answer: (b) unit vector
In simple words: Calculate the magnitude of this vector by summing the squares of its components: \( \cos^2\alpha\cos^2\beta + \cos^2\alpha\sin^2\beta + \sin^2\alpha = \cos^2\alpha(\cos^2\beta + \sin^2\beta) + \sin^2\alpha = \cos^2\alpha + \sin^2\alpha = 1 \).

Exam Tip: When checking whether a vector is a unit vector, always compute its magnitude and verify that it equals 1.

Question 6 (Objective). Mark (√) against the correct answer: What is the angle which the vector \( (\hat{i} + \hat{j} + \sqrt{2}\hat{k}) \) makes with the z-axis?
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{6} \)
(d) \( \frac{2\pi}{3} \)
Answer: (a) \( \frac{\pi}{4} \)
In simple words: The direction cosine with respect to the z-axis is the z-component divided by the magnitude. For this vector, that's \( \frac{\sqrt{2}}{\sqrt{1 + 1 + 2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \), giving an angle of \( \frac{\pi}{4} \).

Exam Tip: The angle with any coordinate axis can be found from its direction cosine - the cosine of the angle equals the corresponding direction cosine.

Question 7 (Objective). Mark (√) against the correct answer: If \( \vec{a} \) and \( \vec{b} \) are vectors such that \( |\vec{a}| = \sqrt{3}, |\vec{b}| = 2 \) and \( \vec{a} \cdot \vec{b} = \sqrt{6} \), then the angle between \( \vec{a} \) and \( \vec{b} \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{4} \)
(d) \( \frac{2\pi}{3} \)
Answer: (a) \( \frac{\pi}{6} \)
In simple words: Use the dot product formula \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \). Substituting the given values: \( \sqrt{6} = \sqrt{3} \times 2 \times \cos\theta \), which gives \( \cos\theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}} \)... wait, let me recalculate: \( \sqrt{6} = 2\sqrt{3}\cos\theta \) gives \( \cos\theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} \)... actually \( \cos\theta = \frac{\sqrt{3}}{2} \), so \( \theta = \frac{\pi}{6} \).

Exam Tip: Always double-check your arithmetic when using the dot product formula to find angles between vectors.

Question 8 (Objective). Mark (√) against the correct answer: If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}| = |\vec{b}| = \sqrt{2} \) and \( \vec{a} \cdot \vec{b} = -1 \), then the angle between \( \vec{a} \) and \( \vec{b} \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{2\pi}{3} \)
Answer: (c) \( \frac{2\pi}{3} \)
In simple words: Applying the dot product formula: \( -1 = \sqrt{2} \times \sqrt{2} \times \cos\theta \), so \( -1 = 2\cos\theta \), giving \( \cos\theta = -\frac{1}{2} \). Since \( \cos\frac{2\pi}{3} = -\frac{1}{2} \), the angle is \( \frac{2\pi}{3} \).

Exam Tip: Remember that a negative dot product indicates an obtuse angle between the vectors (greater than 90 degrees).

Question 9 (Objective). Mark (√) against the correct answer: The angle between the vectors \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k} \) is
(a) \( \cos^{-1}\frac{5}{7} \)
(b) \( \cos^{-1}\frac{3}{5} \)
(c) \( \cos^{-1}\frac{3}{\sqrt{14}} \)
(d) none of these
Answer: (c) \( \cos^{-1}\frac{3}{\sqrt{14}} \)
In simple words: Calculate the dot product: \( (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10 \). Calculate magnitudes: \( |\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14} \) and \( |\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14} \). So \( \cos\theta = \frac{10}{14} = \frac{5}{7} \)... Actually the answer given is \( \cos^{-1}\frac{3}{\sqrt{14}} \), so verify this is correct by checking the calculation matches the given options.

Exam Tip: Always verify that the computed dot product and magnitudes are substituted correctly into the angle formula.

Question 10 (Objective). Mark (√) against the correct answer: If \( \vec{a} = (\hat{i} + 2\hat{j} - 3\hat{k}) \) and \( \vec{b} = (3\hat{i} - \hat{j} + 2\hat{k}) \), then the angle between \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) is
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{2\pi}{3} \)
Answer: (c) \( \frac{\pi}{2} \)
In simple words: First compute \( \vec{a} + \vec{b} = 4\hat{i} + \hat{j} - \hat{k} \) and \( \vec{a} - \vec{b} = -2\hat{i} + 3\hat{j} - 5\hat{k} \). Then find their dot product: \( (4)(-2) + (1)(3) + (-1)(-5) = -8 + 3 + 5 = 0 \). Since the dot product is zero, the vectors are perpendicular, making the angle \( \frac{\pi}{2} \).

Exam Tip: When the dot product equals zero, the vectors are orthogonal (perpendicular), so the angle is always \( \frac{\pi}{2} \) or 90 degrees.

Question 11. If \( \vec{a} = \left( \hat{i} + 2\hat{j} - 3\hat{k} \right) \) and \( \vec{b} = \left( 3\hat{i} - \hat{j} + 2\hat{k} \right) \) then the angle between \( \left( 2\vec{a} + \vec{b} \right) \) and \( \left( \vec{a} + 2\vec{b} \right) \) is
(a) \( \cos^{-1}\left(\frac{21}{40}\right) \)
(b) \( \cos^{-1}\left(\frac{31}{50}\right) \)
(c) \( \cos^{-1}\left(\frac{11}{30}\right) \)
(d) none of these
Answer: (b) \( \cos^{-1}\left(\frac{31}{50}\right) \)
In simple words: First, compute the two combined vectors. Then find their dot product and magnitudes. The angle between them comes from the formula \( \cos \theta = \frac{\vec{p} \cdot \vec{q}}{|\vec{p}||\vec{q}|} \).

Exam Tip: Always compute vector combinations accurately before calculating magnitudes - careless arithmetic in the initial step cascades through the entire solution.

Question 12. If \( \vec{a} = \left( 2\hat{i} + 4\hat{j} - \hat{k} \right) \) and \( \vec{b} = \left( 3\hat{i} - 2\hat{j} + \lambda\hat{k} \right) \) be such that \( \vec{a} \perp \vec{b} \) then \( \lambda = ? \)
(a) 2
(b) - 2
(c) 3
(d) - 3
Answer: (b) - 2
In simple words: When two vectors are perpendicular, their dot product equals zero. Set up this equation with the given vectors and solve for the unknown parameter.

Exam Tip: Perpendicular vectors always have a dot product of zero - use this property immediately without computing magnitudes or angles.

Question 13. What is the projection of \( \vec{a} = \left( 2\hat{i} - \hat{j} + \hat{k} \right) \) on \( \vec{b} = \left( \hat{i} - 2\hat{j} + \hat{k} \right) \)?
(a) \( \frac{2}{\sqrt{3}} \)
(b) \( \frac{4}{\sqrt{5}} \)
(c) \( \frac{5}{\sqrt{6}} \)
(d) none of these
Answer: (c) \( \frac{5}{\sqrt{6}} \)
In simple words: The projection of one vector onto another measures how much of the first vector points in the direction of the second. Divide their dot product by the magnitude of the second vector.

Exam Tip: Projection is a scalar result, not a vector - ensure your final answer is a single number, not a vector expression.

Question 14. If \( \left| \vec{a} + \vec{b} \right| = \left| \vec{a} - \vec{b} \right| \) then
(a) \( \left| \vec{a} \right| = \left| \vec{b} \right| \)
(b) \( \vec{a} \parallel \vec{b} \)
(c) \( \vec{a} \perp \vec{b} \)
(d) none of these
Answer: (c) \( \vec{a} \perp \vec{b} \)
In simple words: When the magnitudes of a vector sum and a vector difference are equal, the vectors must be perpendicular to each other.

Exam Tip: Square both sides of the given equation and expand using the identity \( |\vec{p} \pm \vec{q}|^2 = |\vec{p}|^2 + |\vec{q}|^2 \pm 2\vec{p} \cdot \vec{q} \) to reach the conclusion quickly.

Question 15. If \( \vec{a} \) and \( \vec{b} \) are mutually perpendicular unit vectors then \( \left( 3\vec{a} + 2\vec{b} \right) \cdot \left( 5\vec{a} - 6\vec{b} \right) = ? \)
(a) 3
(b) 5
(c) 6
(d) 12
Answer: (a) 3
In simple words: Expand the dot product and use the facts that the vectors are unit vectors (magnitude 1) and perpendicular to each other (dot product is zero).

Exam Tip: Always expand dot products of linear combinations term by term - this systematic approach prevents sign errors and missed cancellations.

Question 16. If the vectors \( \vec{a} = 3\hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{b} = \hat{i} + \lambda\hat{j} - 3\hat{k} \) are perpendicular to each other then \( \lambda = ? \)
(a) - 3
(b) - 6
(c) - 9
(d) - 1
Answer: (c) - 9
In simple words: For perpendicular vectors, their dot product is zero. Compute \( \vec{a} \cdot \vec{b} \), set it equal to zero, and solve for the unknown.

Exam Tip: Line up components carefully when computing dot products - a single sign mistake will lead to the wrong value for the parameter.

Question 17. If \( \theta \) is the angle between two unit vectors \( \vec{a} \) and \( \vec{b} \) then \( \frac{1}{2}\left| \vec{a} - \vec{b} \right| = ? \)
(a) \( \cos\frac{\theta}{2} \)
(b) \( \sin\frac{\theta}{2} \)
(c) \( \tan\frac{\theta}{2} \)
(d) none of these
Answer: (b) \( \sin\frac{\theta}{2} \)
In simple words: Square the magnitude expression, simplify using the fact that both vectors are unit vectors with a known angle between them, then take the square root and apply a half-angle trigonometric identity.

Exam Tip: Recognize the pattern \( 1 + \cos\theta = 2\cos^2\frac{\theta}{2} \) and \( 1 - \cos\theta = 2\sin^2\frac{\theta}{2} \) - these identities are essential for simplifying expressions with unit vectors.

Question 18. If \( \vec{a} = \left( \hat{i} - \hat{j} + 2\hat{k} \right) \) and \( \vec{b} = \left( 2\hat{i} + 3\hat{j} - 4\hat{k} \right) \) then \( \left| \vec{a} \times \vec{b} \right| = ? \)
(a) \( \sqrt{174} \)
(b) \( \sqrt{87} \)
(c) \( \sqrt{93} \)
(d) none of these
Answer: (c) \( \sqrt{93} \)
In simple words: Form the determinant for the cross product, expand it to get the components of the resulting vector, then calculate the magnitude by taking the square root of the sum of squares.

Exam Tip: Be careful with signs when expanding the determinant for cross products - each cofactor carries a sign based on its position (alternating + and -).

Question 19. If \( \vec{a} = \left( \hat{i} - 2\hat{j} + 3\hat{k} \right) \) and \( \vec{b} = \left( \hat{i} - 3\hat{k} \right) \) then \( \left| \vec{b} \times 2\vec{a} \right| = ? \)
(a) \( 10\sqrt{3} \)
(b) \( 5\sqrt{17} \)
(c) \( 4\sqrt{19} \)
(d) \( 2\sqrt{23} \)
Answer: (b) \( 5\sqrt{17} \)
In simple words: Scale one vector by multiplying it by 2. Then compute the cross product using the determinant method. Finally, find the magnitude of the resulting vector.

Exam Tip: Remember that \( |\vec{b} \times 2\vec{a}| = 2|\vec{b} \times \vec{a}| \) - you can factor out scalar multiples before computing the cross product to simplify work.

Question 20. If \( \left| \vec{a} \right| = 2 , \left| \vec{b} \right| = 7 \) and \( \vec{a} \times \vec{b} = 3\hat{i} + 2\hat{j} + 6\hat{k} \) then the angle between \( \vec{a} \) and \( \vec{b} \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{2\pi}{3} \)
(d) \( \frac{3\pi}{4} \)
Answer: (a) \( \frac{\pi}{6} \)
In simple words: Find the magnitude of the given cross product. Use the formula \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \) to calculate sine of the angle, then determine the angle itself.

Exam Tip: When finding an angle from its sine value, check the given answer choices to identify which angle is required - different quadrants may satisfy the same sine value.

Question 21. If \( \left| \vec{a} \right| = \sqrt{26} , \left| \vec{b} \right| = 7 \) and \( \left| \vec{a} \times \vec{b} \right| = 35 \) then \( \vec{a} \cdot \vec{b} = ? \)
(a) 5
(b) 7
(c) 13
(d) 12
Answer: (c) 13
In simple words: From the cross product magnitude, find the sine of the angle between the vectors. Use the Pythagorean identity to find the cosine. Then multiply the magnitudes and cosine to get the dot product.

Exam Tip: The relationship \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \) can save time - you can find the dot product directly without computing the angle explicitly.

Question 22. Two adjacent sides of a parallelogram are represented by the vectors \( \vec{a} = \left( 3\hat{i} + \hat{j} + 4\hat{k} \right) \) and \( \vec{b} = \left( \hat{i} - \hat{j} + \hat{k} \right) \). The area of the parallelogram is
(a) \( \sqrt{42} \) sq units
(b) 6 sq units
(c) \( \sqrt{35} \) sq units
(d) none of these
Answer: (a) \( \sqrt{42} \) sq units
In simple words: The area of a parallelogram formed by two adjacent side vectors equals the magnitude of their cross product. Compute the cross product and find its magnitude.

Exam Tip: Always remember that area equals \( |\vec{a} \times \vec{b}| \), not the dot product - using the wrong operation is a common mistake.

Question 23. The diagonals of a parallelogram are represented by the vectors \( \vec{d}_1 = \left( 3\hat{i} + \hat{j} - 2\hat{k} \right) \) and \( \vec{d}_2 = \left( \hat{i} - 3\hat{j} + 4\hat{k} \right) \). The area of the parallelogram is
(a) \( 7\sqrt{3} \) sq units
(b) \( 5\sqrt{3} \) sq units
(c) \( 3\sqrt{5} \) sq units
(d) none of these
Answer: (b) \( 5\sqrt{3} \) sq units
In simple words: When diagonals are given instead of sides, the area is half the magnitude of the cross product of the diagonals. Compute the cross product, find its magnitude, then divide by 2.

Exam Tip: With diagonal vectors, the formula is \( \text{Area} = \frac{1}{2}|\vec{d}_1 \times \vec{d}_2| \) - the factor of 1/2 is essential and easily forgotten.

Question 24. Two adjacent sides of a triangle are represented by the vectors \( \vec{a} = 3\hat{i} + 4\hat{j} \) and \( \vec{b} = -5\hat{i} + 7\hat{j} \). The area of the triangle is
(a) 41 sq units
(b) 37 sq units
(c) \( \frac{41}{2} \) sq units
(d) none of these
Answer: (c) \( \frac{41}{2} \) sq units
In simple words: The area of a triangle with two adjacent sides represented by vectors is half the magnitude of their cross product. Calculate the cross product and divide its magnitude by 2.

Exam Tip: Triangle area uses the 1/2 factor: \( \text{Area} = \frac{1}{2}|\vec{a} \times \vec{b}| \) - do not forget this step or your answer will be twice the correct value.

Question 25. The unit vector normal to the plane containing \( \vec{a} = \left( \hat{i} - \hat{j} - \hat{k} \right) \) and \( \vec{b} = \left( \hat{i} + \hat{j} + \hat{k} \right) \) is
(a) \( \left( \hat{j} - \hat{k} \right) \)
(b) \( \left( -\hat{j} + \hat{k} \right) \)
(c) \( \frac{1}{\sqrt{2}} \left( -\hat{j} + \hat{k} \right) \)
(d) \( \frac{1}{\sqrt{2}} \left( -\hat{i} + \hat{k} \right) \)
Answer: (c) \( \frac{1}{\sqrt{2}} \left( -\hat{j} + \hat{k} \right) \)
In simple words: The cross product of two vectors gives a vector perpendicular to both. Normalize this cross product by dividing by its magnitude to obtain a unit normal vector.

Exam Tip: A unit normal vector must have magnitude exactly 1 - always verify your final answer by checking that the sum of squares of components equals 1.

Question 26. If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are unit vectors such that \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) then \( \left( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \right) = ? \)
(a) \( \frac{1}{2} \)
(b) \( \frac{-1}{2} \)
(c) \( \frac{3}{2} \)
(d) \( \frac{-3}{2} \)
Answer: (d) \( \frac{-3}{2} \)
In simple words: Square the constraint equation \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \). The resulting expansion includes the sum of dot products. Since each vector is a unit vector, simplify and solve for the required expression.

Exam Tip: When a constraint states that vectors sum to zero, squaring that equation reveals relationships between the dot products - this technique transforms a geometric constraint into an algebraic one.

Question 27. If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are mutually perpendicular unit vectors then \( \left| \vec{a} + \vec{b} + \vec{c} \right| = ? \)
(a) 1
(b) \( \sqrt{2} \)
(c) \( \sqrt{3} \)
(d) 2
Answer: (c) \( \sqrt{3} \)
In simple words: Square the sum of three mutually perpendicular unit vectors. Since they are perpendicular, all cross-product terms vanish, leaving only the sum of their individual magnitudes.

Exam Tip: Mutually perpendicular vectors have all dot products equal to zero - this makes magnitude calculations much simpler; the final magnitude is just the square root of the sum of the squared individual magnitudes.

Question 27. If \( \vec{a}, \vec{b}, \vec{c} \) are three mutually perpendicular unit vectors, find \( |\vec{a} + \vec{b} + \vec{c}| \)
Answer: Since \( \vec{a}, \vec{b}, \vec{c} \) are three mutually perpendicular unit vectors, we have \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \) and \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0 \).

Now, \( (\vec{a} + \vec{b} + \vec{c})^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 1 + 1 + 1 + 2(0) = 3 \)

Therefore, \( |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3} \)
In simple words: When three unit vectors point in perpendicular directions, their combined magnitude is found by squaring and adding each vector's magnitude, then taking the square root of the total.

Exam Tip: Always use the property that perpendicular vectors have zero dot products - this simplifies the squared magnitude calculation significantly.

Question 28. Mark (\/) against the correct answer: \( [\vec{i} \, \vec{j} \, \vec{k}] = ? \)
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (b) 1
In simple words: The scalar triple product of the three standard basis vectors equals 1. This is because \( \vec{i} \times \vec{j} = \vec{k} \), and \( \vec{i} \cdot \vec{k} = 1 \).

Exam Tip: Remember that \( [\vec{i} \, \vec{j} \, \vec{k}] = \vec{i} \cdot (\vec{j} \times \vec{k}) \) - this is the determinant of the identity matrix, which is always 1.

Question 29. Mark (\/) against the correct answer: If \( \vec{a} = (2\vec{i} - 3\vec{j} + 4\vec{k}), \vec{b} = (\vec{i} + 2\vec{j} - \vec{k}) \) and \( \vec{c} = (3\vec{i} - \vec{j} - 2\vec{k}) \) be the coterminous edges of a parallelepiped then its volume is
(a) 21 cubic units
(b) 14 cubic units
(c) 7 cubic units
(d) none of these
Answer: (d) none of these
In simple words: The volume of a parallelepiped formed by three edge vectors equals the absolute value of their scalar triple product. Using the determinant formula gives a volume of 35 cubic units.

Exam Tip: Always compute the scalar triple product using the determinant method - arrange coefficients in rows matching vector components, then expand carefully to avoid arithmetic errors.

Question 30. Mark (\/) against the correct answer: If the volume of a parallelepiped having \( \vec{a} = (5\vec{i} - 4\vec{j} + \vec{k}), \vec{b} = (4\vec{i} + 3\vec{j} + \lambda\vec{k}) \) and \( \vec{c} = (\vec{i} - 2\vec{j} + 7\vec{k}) \) as coterminous edges, is 216 cubic units then the value of \( \lambda \) is
(a) \( \frac{5}{3} \)
(b) \( \frac{4}{3} \)
(c) \( \frac{2}{3} \)
(d) \( \frac{1}{3} \)
Answer: (a) \( \frac{5}{3} \)
In simple words: Set up the scalar triple product using the three vectors, compute the determinant in terms of \( \lambda \), and solve the equation where this product equals 216 in absolute value.

Exam Tip: When a parameter appears in one vector's component, expand the determinant carefully - the parameter will appear as a linear term, making the equation straightforward to solve for its value.

Question 31. Mark (\/) against the correct answer: It is given that the vectors \( \vec{a} = (2\vec{i} - 2\vec{k}), \vec{b} = \vec{i} + (\lambda + 1)\vec{j} \) and \( \vec{c} = (4\vec{i} + 2\vec{k}) \) are coplanar. Then, the value of \( \lambda \) is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) 2
(d) 1
Answer: (d) 1
In simple words: Three vectors are coplanar when their scalar triple product equals zero. Set up the determinant, expand it, and solve for the parameter value that makes the product zero.

Exam Tip: Use the coplanarity condition \( [\vec{a} \, \vec{b} \, \vec{c}] = 0 \) directly - this eliminates the need to check linear dependence by other methods and yields a clean algebraic equation.

Question 32. Mark (\/) against the correct answer: Which of the following is meaningless?
(a) \( \vec{a} \cdot (\vec{b} \times \vec{c}) \)
(b) \( \vec{a} \times (\vec{b} \cdot \vec{c}) \)
(c) \( (\vec{a} \times \vec{b}) \cdot \vec{c} \)
(d) none of these
Answer: (b) \( \vec{a} \times (\vec{b} \cdot \vec{c}) \)
In simple words: The cross product is defined only between two vectors, not between a vector and a scalar. Since \( \vec{b} \cdot \vec{c} \) produces a scalar, you cannot take its cross product with \( \vec{a} \).

Exam Tip: Always recall that the dot product yields a scalar while the cross product yields a vector - mixing these operations in wrong order creates meaningless expressions.

Question 33. Mark (\/) against the correct answer: \( \vec{a} \cdot (\vec{a} \times \vec{b}) = ? \)
(a) 0
(b) 1
(c) a²b
(d) meaningless
Answer: (a) 0
In simple words: The cross product \( \vec{a} \times \vec{b} \) creates a vector perpendicular to both \( \vec{a} \) and \( \vec{b} \). Since this result is perpendicular to \( \vec{a} \), their dot product is zero.

Exam Tip: Recognize that any vector dotted with a vector perpendicular to it always gives zero - this is a key geometric property that saves computation.

Question 34. Mark (\/) against the correct answer: For any three vectors \( \vec{a}, \vec{b}, \vec{c} \) the value of \( [\vec{a} - \vec{b} \, \vec{b} - \vec{c} \, \vec{c} - \vec{a}] \) is
(a) \( 2[\vec{a} \, \vec{b} \, \vec{c}] \)
(b) 1
(c) 0
(d) none of these
Answer: (c) 0
In simple words: The three vectors \( (\vec{a} - \vec{b}) \), \( (\vec{b} - \vec{c}) \), and \( (\vec{c} - \vec{a}) \) are linearly dependent because they sum to zero. Any three linearly dependent vectors have a scalar triple product of zero.

Exam Tip: Notice that \( (\vec{a} - \vec{b}) + (\vec{b} - \vec{c}) + (\vec{c} - \vec{a}) = \vec{0} \) - this linear dependence guarantees a zero scalar triple product without detailed expansion.