RS Aggarwal Solutions for Class 12 Chapter 23 Scalar, or Dot, Product of Vectors

Question 1. Find \( \overrightarrow{a} \cdot \overrightarrow{b} \) when
(i) \( \overrightarrow{a} = \overrightarrow{i} - 2\overrightarrow{j} + \overrightarrow{k} \) and \( \overrightarrow{b} = 3\overrightarrow{i} - 4\overrightarrow{j} - 2\overrightarrow{k} \)
(ii) \( \overrightarrow{a} = \overrightarrow{i} + 2\overrightarrow{j} + 3\overrightarrow{k} \) and \( \overrightarrow{b} = -2\overrightarrow{j} + 4\overrightarrow{k} \)
(iii) \( \overrightarrow{a} = \overrightarrow{i} - \overrightarrow{j} + 5\overrightarrow{k} \) and \( \overrightarrow{b} = 3\overrightarrow{i} + 0\overrightarrow{j} - 2\overrightarrow{k} \)
Answer:
(i) Starting with the given vectors, we apply the dot product formula by multiplying corresponding components and summing them:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (1)(3) + (-2)(-4) + (1)(-2) = 3 + 8 - 2 = 9 \)
(ii) Here the vectors are \( \overrightarrow{a} = \overrightarrow{i} + 2\overrightarrow{j} + 3\overrightarrow{k} \) and \( \overrightarrow{b} = 0\overrightarrow{i} - 2\overrightarrow{j} + 4\overrightarrow{k} \). Computing component-wise:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (1)(0) + (2)(-2) + (3)(4) = 0 - 4 + 12 = 8 \)
(iii) With \( \overrightarrow{a} = \overrightarrow{i} - \overrightarrow{j} + 5\overrightarrow{k} \) and \( \overrightarrow{b} = 3\overrightarrow{i} + 0\overrightarrow{j} - 2\overrightarrow{k} \), the calculation gives:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (1)(3) + (-1)(0) + (5)(-2) = 3 + 0 - 10 = -7 \)
In simple words: Multiply each component of the first vector by the matching component of the second vector, then add all three results together to get the final answer.

Exam Tip: Always align components by their direction (i-component with i-component, j with j, k with k) before multiplying. This prevents sign errors and ensures accuracy.

Question 2. Find the value of λ for which \( \overrightarrow{a} \) and \( \overrightarrow{b} \) are perpendicular, where
(i) \( \overrightarrow{a} = 2\overrightarrow{i} + \lambda\overrightarrow{j} + \overrightarrow{k} \) and \( \overrightarrow{b} = (\overrightarrow{i} - 2\overrightarrow{j} + 3\overrightarrow{k}) \)
(ii) \( \overrightarrow{a} = 3\overrightarrow{i} - \overrightarrow{j} + 4\overrightarrow{k} \) and \( \overrightarrow{b} = -\lambda\overrightarrow{i} + 3\overrightarrow{j} + 3\overrightarrow{k} \)
(iii) \( \overrightarrow{a} = 2\overrightarrow{i} + 4\overrightarrow{j} - \overrightarrow{k} \) and \( \overrightarrow{b} = 3\overrightarrow{i} - 2\overrightarrow{j} + \lambda\overrightarrow{k} \)
(iv) \( \overrightarrow{a} = 3\overrightarrow{i} + 2\overrightarrow{j} - 5\overrightarrow{k} \) and \( \overrightarrow{b} = -5\overrightarrow{j} + \lambda\overrightarrow{k} \)
Answer:
(i) Two vectors are perpendicular when their dot product equals zero. Setting up the equation:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (2)(1) + (\lambda)(-2) + (1)(3) = 0 \)
\( 2 - 2\lambda + 3 = 0 \)
\( 5 = 2\lambda \)
\( \lambda = \frac{5}{2} \)

(ii) For perpendicularity, the dot product must be zero:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (3)(-\lambda) + (-1)(3) + (4)(3) = 0 \)
\( -3\lambda - 3 + 12 = 0 \)
\( -3\lambda + 9 = 0 \)
\( \lambda = 3 \)

(iii) Applying the perpendicularity condition:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (2)(3) + (4)(-2) + (-1)(\lambda) = 0 \)
\( 6 - 8 - \lambda = 0 \)
\( -\lambda - 2 = 0 \)
\( \lambda = -2 \)

(iv) Setting the dot product to zero:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (3)(0) + (2)(-5) + (-5)(\lambda) = 0 \)
\( 0 - 10 - 5\lambda = 0 \)
\( -5\lambda = 10 \)
\( \lambda = -2 \)
In simple words: Perpendicular vectors have a dot product of zero. Multiply the components and set the sum equal to zero, then solve for λ.

Exam Tip: Remember that perpendicularity means the dot product equals exactly zero - this is the key condition. Always double-check your arithmetic when solving for λ.

Question 3. (i) If \( \overrightarrow{a} = \overrightarrow{i} + 2\overrightarrow{j} - 3\overrightarrow{k} \) and \( \overrightarrow{b} = 3\overrightarrow{i} - \overrightarrow{j} + 2\overrightarrow{k} \), show that \( (\overrightarrow{a} + \overrightarrow{b}) \) is perpendicular to \( (\overrightarrow{a} - \overrightarrow{b}) \).
Answer: We first compute the sum and difference of the vectors:
\( \overrightarrow{a} + \overrightarrow{b} = (\overrightarrow{i} + 2\overrightarrow{j} - 3\overrightarrow{k}) + (3\overrightarrow{i} - \overrightarrow{j} + 2\overrightarrow{k}) = 4\overrightarrow{i} + \overrightarrow{j} - \overrightarrow{k} \)
\( \overrightarrow{a} - \overrightarrow{b} = (\overrightarrow{i} + 2\overrightarrow{j} - 3\overrightarrow{k}) - (3\overrightarrow{i} - \overrightarrow{j} + 2\overrightarrow{k}) = -2\overrightarrow{i} + 3\overrightarrow{j} - 5\overrightarrow{k} \)

Now we calculate their dot product:
\( (\overrightarrow{a} + \overrightarrow{b}) \cdot (\overrightarrow{a} - \overrightarrow{b}) = (4)(-2) + (1)(3) + (-1)(-5) = -8 + 3 + 5 = 0 \)

Since the dot product equals zero, the vectors are perpendicular.
In simple words: Add the two vectors and subtract them to get two new vectors. Then multiply their corresponding parts and add - if you get zero, they are perpendicular.

Exam Tip: This is a standard proof question. Show all three steps clearly: find the sum, find the difference, then compute the dot product and conclude.

Question 3. (ii) If \( \overrightarrow{a} = (5\overrightarrow{i} - \overrightarrow{j} - 3\overrightarrow{k}) \) and \( \overrightarrow{b} = (\overrightarrow{i} + 3\overrightarrow{j} - 5\overrightarrow{k}) \), then show that \( (\overrightarrow{a} + \overrightarrow{b}) \) and \( (\overrightarrow{a} - \overrightarrow{b}) \) are orthogonal.
Answer: We find the sum and difference:
\( \overrightarrow{a} + \overrightarrow{b} = (5\overrightarrow{i} - \overrightarrow{j} - 3\overrightarrow{k}) + (\overrightarrow{i} + 3\overrightarrow{j} - 5\overrightarrow{k}) = 6\overrightarrow{i} + 2\overrightarrow{j} - 8\overrightarrow{k} \)
\( \overrightarrow{a} - \overrightarrow{b} = (5\overrightarrow{i} - \overrightarrow{j} - 3\overrightarrow{k}) - (\overrightarrow{i} + 3\overrightarrow{j} - 5\overrightarrow{k}) = 4\overrightarrow{i} - 4\overrightarrow{j} + 2\overrightarrow{k} \)

Computing the dot product to check orthogonality:
\( (\overrightarrow{a} + \overrightarrow{b}) \cdot (\overrightarrow{a} - \overrightarrow{b}) = (6)(4) + (2)(-4) + (-8)(2) = 24 - 8 - 16 = 0 \)

The dot product equals zero, confirming that the vectors are orthogonal (perpendicular).
In simple words: Two vectors are orthogonal when their dot product is zero. Add and subtract the vectors, multiply matching components, and if the total is zero, they are orthogonal.

Exam Tip: Orthogonal is another word for perpendicular. Always verify by computing the dot product explicitly rather than assuming.

Question 4. If \( \overrightarrow{a} = (\overrightarrow{i} - \overrightarrow{j} + 7\overrightarrow{k}) \) and \( \overrightarrow{b} = (5\overrightarrow{i} - \overrightarrow{j} + \lambda\overrightarrow{k}) \) then find the value of λ so that \( (\overrightarrow{a} + \overrightarrow{b}) \) and \( (\overrightarrow{a} - \overrightarrow{b}) \) are orthogonal vectors.
Answer: First, we compute the sum and difference:
\( \overrightarrow{a} + \overrightarrow{b} = (\overrightarrow{i} - \overrightarrow{j} + 7\overrightarrow{k}) + (5\overrightarrow{i} - \overrightarrow{j} + \lambda\overrightarrow{k}) = 6\overrightarrow{i} - 2\overrightarrow{j} + (7 + \lambda)\overrightarrow{k} \)
\( \overrightarrow{a} - \overrightarrow{b} = (\overrightarrow{i} - \overrightarrow{j} + 7\overrightarrow{k}) - (5\overrightarrow{i} - \overrightarrow{j} + \lambda\overrightarrow{k}) = -4\overrightarrow{i} + 0\overrightarrow{j} + (7 - \lambda)\overrightarrow{k} \)

For orthogonality, set the dot product to zero:
\( (\overrightarrow{a} + \overrightarrow{b}) \cdot (\overrightarrow{a} - \overrightarrow{b}) = (6)(-4) + (-2)(0) + (7 + \lambda)(7 - \lambda) = 0 \)
\( -24 + 0 + (49 - \lambda^2) = 0 \)
\( -24 + 49 - \lambda^2 = 0 \)
\( 25 - \lambda^2 = 0 \)
\( \lambda^2 = 25 \)
\( \lambda = \pm 5 \)
In simple words: Build the sum and difference vectors, then require their dot product to equal zero. This gives a condition on λ that you can solve to find its value.

Exam Tip: When a squared term appears, remember both positive and negative roots are valid unless the problem specifies otherwise. Verify both solutions if there is time.

Question 5. Show that the vectors \( \overrightarrow{a} = \frac{1}{7}(2\overrightarrow{i} + 3\overrightarrow{j} + 6\overrightarrow{k}) \), \( \overrightarrow{b} = \frac{1}{7}(3\overrightarrow{i} - 6\overrightarrow{j} + 2\overrightarrow{k}) \) and \( \overrightarrow{c} = \frac{1}{7}(6\overrightarrow{i} + 2\overrightarrow{j} - 3\overrightarrow{k}) \) are mutually perpendicular unit vectors.
Answer: To show mutual perpendicularity, we verify that each pair has a dot product of zero.

First, check perpendicularity of \( \overrightarrow{a} \) and \( \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = \frac{1}{49}[(2)(3) + (3)(-6) + (6)(2)] = \frac{1}{49}(6 - 18 + 12) = \frac{1}{49}(0) = 0 \)

Next, check perpendicularity of \( \overrightarrow{b} \) and \( \overrightarrow{c} \):
\( \overrightarrow{b} \cdot \overrightarrow{c} = \frac{1}{49}[(3)(6) + (-6)(2) + (2)(-3)] = \frac{1}{49}(18 - 12 - 6) = \frac{1}{49}(0) = 0 \)

Finally, check perpendicularity of \( \overrightarrow{a} \) and \( \overrightarrow{c} \):
\( \overrightarrow{a} \cdot \overrightarrow{c} = \frac{1}{49}[(2)(6) + (3)(2) + (6)(-3)] = \frac{1}{49}(12 + 6 - 18) = \frac{1}{49}(0) = 0 \)

To show they are unit vectors, calculate the magnitude of each:
\( |\overrightarrow{a}| = \frac{1}{7}\sqrt{2^2 + 3^2 + 6^2} = \frac{1}{7}\sqrt{4 + 9 + 36} = \frac{1}{7}\sqrt{49} = \frac{7}{7} = 1 \)
\( |\overrightarrow{b}| = \frac{1}{7}\sqrt{3^2 + (-6)^2 + 2^2} = \frac{1}{7}\sqrt{9 + 36 + 4} = \frac{1}{7}\sqrt{49} = 1 \)
\( |\overrightarrow{c}| = \frac{1}{7}\sqrt{6^2 + 2^2 + (-3)^2} = \frac{1}{7}\sqrt{36 + 4 + 9} = \frac{1}{7}\sqrt{49} = 1 \)

Since all three pairwise dot products are zero and each has magnitude 1, the vectors are mutually perpendicular unit vectors.
In simple words: Show that each pair of vectors multiplies to zero (meaning they are perpendicular), and that each vector has length 1 (meaning they are unit vectors).

Exam Tip: When proving three vectors are mutually perpendicular, you must check all three pairs. Do not skip any pairing, as examiners specifically look for completeness.

Question 6. Let \( \overrightarrow{a} = 4\overrightarrow{i} + 5\overrightarrow{j} - \overrightarrow{k} \), \( \overrightarrow{b} = \overrightarrow{i} - 4\overrightarrow{j} + 5\overrightarrow{k} \) and \( \overrightarrow{c} = 3\overrightarrow{i} + \overrightarrow{j} - \overrightarrow{k} \). Find a vector \( \overrightarrow{d} \) which is perpendicular to both \( \overrightarrow{a} \) and \( \overrightarrow{b} \), and is such that \( \overrightarrow{d} \cdot \overrightarrow{c} = 21 \).
Answer: Let \( \overrightarrow{d} = p\overrightarrow{i} + q\overrightarrow{j} + r\overrightarrow{k} \). Since \( \overrightarrow{d} \) is perpendicular to both \( \overrightarrow{a} \) and \( \overrightarrow{b} \), we have:
\( \overrightarrow{d} \cdot \overrightarrow{a} = 0 \Rightarrow (p\overrightarrow{i} + q\overrightarrow{j} + r\overrightarrow{k}) \cdot (4\overrightarrow{i} + 5\overrightarrow{j} - \overrightarrow{k}) = 0 \)
\( 4p + 5q - r = 0 \quad \ldots \text{(i)} \)

\( \overrightarrow{d} \cdot \overrightarrow{b} = 0 \Rightarrow (p\overrightarrow{i} + q\overrightarrow{j} + r\overrightarrow{k}) \cdot (\overrightarrow{i} - 4\overrightarrow{j} + 5\overrightarrow{k}) = 0 \)
\( p - 4q + 5r = 0 \quad \ldots \text{(ii)} \)

From the condition \( \overrightarrow{d} \cdot \overrightarrow{c} = 21 \):
\( (p\overrightarrow{i} + q\overrightarrow{j} + r\overrightarrow{k}) \cdot (3\overrightarrow{i} + \overrightarrow{j} - \overrightarrow{k}) = 21 \)
\( 3p + q - r = 21 \quad \ldots \text{(iii)} \)

Solving equations (i), (ii), and (iii) simultaneously:
From equation (i): \( r = 4p + 5q \)

Substitute into equation (ii):
\( p - 4q + 5(4p + 5q) = 0 \)
\( p - 4q + 20p + 25q = 0 \)
\( 21p + 21q = 0 \)
\( p = -q \)

Substitute into equation (iii):
\( 3p + q - (4p + 5q) = 21 \)
\( 3p + q - 4p - 5q = 21 \)
\( -p - 4q = 21 \)

With \( p = -q \):
\( -(-q) - 4q = 21 \)
\( q - 4q = 21 \)
\( -3q = 21 \)
\( q = -7 \)

Therefore \( p = 7 \), and \( r = 4(7) + 5(-7) = 28 - 35 = -7 \)

Thus \( \overrightarrow{d} = 7\overrightarrow{i} - 7\overrightarrow{j} - 7\overrightarrow{k} = 7(\overrightarrow{i} - \overrightarrow{j} - \overrightarrow{k}) \)
In simple words: To find a vector perpendicular to two given vectors, set up two equations using the perpendicularity condition (dot product equals zero). Then apply the third condition to solve for the unknown coefficients.

Exam Tip: When finding a vector satisfying multiple conditions, organize the equations carefully and solve them as a system. Always verify your final answer by checking all three conditions.

Question 7. Let \( \overrightarrow{a} = (2\overrightarrow{i} + 3\overrightarrow{j} + 2\overrightarrow{k}) \) and \( \overrightarrow{b} = (\overrightarrow{i} + 2\overrightarrow{j} + \overrightarrow{k}) \). Find the projection of (i) \( \overrightarrow{a} \) on \( \overrightarrow{b} \) and (ii) \( \overrightarrow{b} \) on \( \overrightarrow{a} \).
Answer: The projection formula is: projection of \( \overrightarrow{a} \) on \( \overrightarrow{b} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|} \)

First, calculate the magnitudes:
\( |\overrightarrow{a}| = \sqrt{2^2 + 3^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \)
\( |\overrightarrow{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \)

(i) Projection of \( \overrightarrow{a} \) on \( \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10 \)
\( \text{Projection} = \frac{10}{\sqrt{6}} = \frac{10\sqrt{6}}{6} = \frac{5\sqrt{6}}{3} \)

(ii) Projection of \( \overrightarrow{b} \) on \( \overrightarrow{a} \):
\( \overrightarrow{b} \cdot \overrightarrow{a} = (1)(2) + (2)(3) + (1)(2) = 2 + 6 + 2 = 10 \)
\( \text{Projection} = \frac{10}{\sqrt{17}} = \frac{10\sqrt{17}}{17} \)
In simple words: To find how much of one vector points in the direction of another, compute their dot product and divide by the length of the vector you are projecting onto.

Exam Tip: The projection is always a scalar (a number), not a vector. Make sure you divide by the magnitude of the correct vector - the one you are projecting onto.

Question 8. Find the projection of \( (8\overrightarrow{i} + \overrightarrow{j}) \) in the direction of \( (\overrightarrow{i} + 2\overrightarrow{j} - 2\overrightarrow{k}) \).
Answer: The projection of a vector onto another is given by: \( \text{proj} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|} \)

Calculate the dot product:
\( (8\overrightarrow{i} + \overrightarrow{j}) \cdot (\overrightarrow{i} + 2\overrightarrow{j} - 2\overrightarrow{k}) = (8)(1) + (1)(2) + (0)(-2) = 8 + 2 + 0 = 10 \)

Calculate the magnitude of the direction vector:
\( |\overrightarrow{b}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \)

Therefore, the projection is:
\( \text{Projection} = \frac{10}{3} \)
In simple words: Multiply the two vectors component by component and add the results. Then divide by the length of the direction vector.

Exam Tip: Always rationalize or simplify the final answer. If the magnitude is a perfect square, use it to get a clean numerical result.

Question 9. Write the projection of vector \( (\overrightarrow{i} + \overrightarrow{j} + \overrightarrow{k}) \) along the vector \( \overrightarrow{j} \).
Answer: The projection formula is: projection \( = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|} \)

Compute the dot product of \( (\overrightarrow{i} + \overrightarrow{j} + \overrightarrow{k}) \) with \( \overrightarrow{j} \):
\( (\overrightarrow{i} + \overrightarrow{j} + \overrightarrow{k}) \cdot \overrightarrow{j} = (1)(0) + (1)(1) + (1)(0) = 1 \)

The magnitude of \( \overrightarrow{j} \) is:
\( |\overrightarrow{j}| = 1 \)

Therefore, the projection is:
\( \text{Projection} = \frac{1}{1} = 1 \)
In simple words: The j-component of the vector is 1, and since j is a unit vector, the projection onto j is simply the j-coefficient of the original vector.

Exam Tip: Projections onto unit vectors (like i, j, k) are particularly straightforward - the answer is just the corresponding component.

Question 10. (i) Find the projection of \( \overrightarrow{a} \) on \( \overrightarrow{b} \) if \( \overrightarrow{a} \cdot \overrightarrow{b} = 8 \) and \( \overrightarrow{b} = (2\overrightarrow{i} + 6\overrightarrow{j} + 3\overrightarrow{k}) \).
Answer: The projection is obtained using: projection of \( \overrightarrow{a} \) on \( \overrightarrow{b} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|} \)

Calculate the magnitude of \( \overrightarrow{b} \):
\( |\overrightarrow{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \)

Thus, the projection is:
\( \text{Projection} = \frac{8}{7} \)
In simple words: You are given the dot product directly, so just divide it by the magnitude of the vector you are projecting onto.

Exam Tip: When the dot product is already provided, use it directly - do not recalculate it from the vectors.

Question 10. (ii) Write the projection of the vector \( (\overrightarrow{i} + \overrightarrow{j}) \) on the vector \( (\overrightarrow{i} - \overrightarrow{j}) \).
Answer: Using the projection formula:

Calculate the dot product:
\( (\overrightarrow{i} + \overrightarrow{j}) \cdot (\overrightarrow{i} - \overrightarrow{j}) = (1)(1) + (1)(-1) = 1 - 1 = 0 \)

Calculate the magnitude:
\( |\overrightarrow{i} - \overrightarrow{j}| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \)

Therefore, the projection is:
\( \text{Projection} = \frac{0}{\sqrt{2}} = 0 \)
In simple words: When the dot product is zero, it means the vectors are perpendicular, so the projection is also zero.

Exam Tip: A zero projection indicates that the two vectors are perpendicular to each other.

Question 11. Find the angle between the vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \), when
(i) \( \overrightarrow{a} = \overrightarrow{i} - 2\overrightarrow{j} + 3\overrightarrow{k} \) and \( \overrightarrow{b} = 3\overrightarrow{i} - 2\overrightarrow{j} + \overrightarrow{k} \)
(ii) \( \overrightarrow{a} = 3\overrightarrow{i} + \overrightarrow{j} + 2\overrightarrow{k} \) and \( \overrightarrow{b} = 2\overrightarrow{i} - 2\overrightarrow{j} + 4\overrightarrow{k} \)
(iii) \( \overrightarrow{a} = \overrightarrow{i} - \overrightarrow{j} \) and \( \overrightarrow{b} = \overrightarrow{j} + \overrightarrow{k} \)
Answer: The angle θ between two vectors is found using: \( \cos\theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|} \)

(i) Calculate magnitudes:
\( |\overrightarrow{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \)
\( |\overrightarrow{b}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \)

Calculate dot product:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10 \)

Find the angle:
\( \cos\theta = \frac{10}{14 \cdot 14} = \frac{10}{196} = \frac{5}{98} \)

Wait, let me recalculate: \( \cos\theta = \frac{10}{\sqrt{14} \cdot \sqrt{14}} = \frac{10}{14} = \frac{5}{7} \)
\( \theta = \cos^{-1}\left(\frac{5}{7}\right) \)

(ii) Calculate magnitudes:
\( |\overrightarrow{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \)
\( |\overrightarrow{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \)

Calculate dot product:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12 \)
\br>Find the angle:
\( \cos\theta = \frac{12}{\sqrt{14} \cdot \sqrt{24}} = \frac{12}{\sqrt{336}} \)
\( \sqrt{336} = \sqrt{16 \cdot 21} = 4\sqrt{21} \)
\( \cos\theta = \frac{12}{4\sqrt{21}} = \frac{3}{\sqrt{21}} = \frac{3\sqrt{21}}{21} = \frac{\sqrt{21}}{7} \)
\( \theta = \cos^{-1}\left(\frac{\sqrt{21}}{7}\right) \)

(iii) Calculate magnitudes:
\( |\overrightarrow{a}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2} \)
\( |\overrightarrow{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \)

Calculate dot product:
\( \overrightarrow{a} \cdot \overrightarrow{b} = (1)(0) + (-1)(1) + (0)(1) = 0 - 1 + 0 = -1 \)

Find the angle:
\( \cos\theta = \frac{-1}{\sqrt{2} \cdot \sqrt{2}} = \frac{-1}{2} \)
\( \theta = \cos^{-1}\left(-\frac{1}{2}\right) = 120° \text{ or } \frac{2\pi}{3} \text{ radians} \)
In simple words: Find the dot product of the two vectors. Then find their magnitudes. Divide the dot product by the product of the magnitudes to get the cosine of the angle.

Exam Tip: Always ensure the angle is in the correct range (0° to 180°). If cosine is negative, the angle is obtuse (between 90° and 180°).

Question 12. If \( \vec{a} = (\hat{i} - \hat{j}) \) and \( \vec{b} = (\hat{j} + \hat{k}) \) then calculate the angle between \( \vec{a} \) and \( \vec{b} \).
Answer: We have \( \vec{a} = (\hat{i} - \hat{j}) \) and \( \vec{b} = (\hat{j} + \hat{k}) \). The magnitude of vector \( \vec{a} \) is \( |\vec{a}| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \). Similarly, the magnitude of vector \( \vec{b} \) is \( |\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2} \). Using the dot product formula: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \). Computing the dot product: \( (\hat{i} - \hat{j}) \cdot (\hat{j} + \hat{k}) = -1 \). Therefore: \( -1 = \sqrt{2} \cdot \sqrt{2} \cdot \cos\theta = 2\cos\theta \), which gives \( \cos\theta = -\frac{1}{2} \). Thus \( \theta = \cos^{-1}(-\frac{1}{2}) = 120° \).
In simple words: Find the magnitude of each vector by taking the square root of the sum of squares of its components. Compute the dot product by multiplying corresponding components and adding them. Use the dot product formula to solve for the angle.

Exam Tip: Always calculate magnitudes correctly and ensure the dot product is computed component-wise before applying the angle formula.

Question 13. If \( \vec{u} \) is a unit vector such that \( (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 8 \), find \( |\vec{x}| \).
Answer: Given that \( \vec{u} \) is a unit vector, we have \( |\vec{u}| = 1 \). We are given \( (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 8 \). Expanding the left side using the distributive property: \( |\vec{x}|^2 - |\vec{a}|^2 = 8 \). Substituting \( |\vec{a}| = 1 \): \( |\vec{x}|^2 - 1 = 8 \), which gives \( |\vec{x}|^2 = 9 \). Therefore \( |\vec{x}| = 3 \).
In simple words: When you multiply out (vector x minus vector a) times (vector x plus vector a), you get the magnitude squared of x minus the magnitude squared of a. Since a is a unit vector with magnitude 1, subtract 1 from 8 to find the magnitude squared of x equals 9, so the magnitude is 3.

Exam Tip: Recognize the difference-of-squares expansion for vectors - it follows the same algebraic pattern as \( (p - q)(p + q) = p^2 - q^2 \).

Question 14. Find the angles which the vector \( \vec{a} = 3\hat{i} - 6\hat{j} + 2\hat{k} \) makes with the coordinate axes.
Answer: The vector is \( \vec{a} = 3\hat{i} - 6\hat{j} + 2\hat{k} \). First, calculate the magnitude: \( |\vec{a}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \). The angle with the x-axis is: \( \alpha = \cos^{-1}\left(\frac{3}{7}\right) \). The angle with the y-axis is: \( \beta = \cos^{-1}\left(\frac{-6}{7}\right) \). The angle with the z-axis is: \( \gamma = \cos^{-1}\left(\frac{2}{7}\right) \).
In simple words: Divide each component of the vector by its magnitude to get the cosine of the angle with each axis. The x-component divided by magnitude gives the cosine of the x-axis angle, and the same for y and z axes.

Exam Tip: The direction cosines are simply the components divided by the vector's magnitude - always calculate the magnitude first to avoid errors.

Question 15. Show that the vector \( \vec{a} = (\hat{i} + \hat{j} + \hat{k}) \) is equally inclined to the coordinate axes.
Answer: The vector is \( \vec{a} = (\hat{i} + \hat{j} + \hat{k}) \). Calculate the magnitude: \( |\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \). The angle with the x-axis: \( \alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \). The angle with the y-axis: \( \beta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \). The angle with the z-axis: \( \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \). Since all three angles are equal to \( \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \), the vector is equally inclined to all coordinate axes.
In simple words: Each component of the vector equals 1, and the magnitude is root 3. When you divide each component by the magnitude, you get the same value for all three axes, so all three angles are identical.

Exam Tip: A vector is equally inclined to coordinate axes when the direction cosines with respect to all three axes are equal.

Question 16. Find a vector \( \vec{u} \) of magnitude \( 5\sqrt{2} \), making an angle \( \frac{\pi}{4} \) with x-axis, \( \frac{\pi}{2} \) with y-axis and an acute angle \( \theta \) with z-axis.
Answer: We are given that \( |\vec{a}| = 5\sqrt{2} \). From the angle conditions: \( l = \cos\alpha = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \), \( m = \cos\beta = \cos\frac{\pi}{2} = 0 \), \( n = \cos\theta \). Using the identity \( l^2 + m^2 + n^2 = 1 \): \( \left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 + n^2 = 1 \), which gives \( \frac{1}{2} + n^2 = 1 \), so \( n^2 = \frac{1}{2} \). Since the angle is acute, \( n = \frac{1}{\sqrt{2}} \). The vector is: \( \vec{a} = |\vec{a}|(l\hat{i} + m\hat{j} + n\hat{k}) = 5\sqrt{2}\left(\frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}\right) = 5(\hat{i} + \hat{k}) \).
In simple words: Use the direction angles to find the direction cosines (the cosines of the angles). These direction cosines must satisfy the identity that their squares sum to 1. Once you have all three direction cosines, multiply the magnitude by each one to get the vector components.

Exam Tip: Remember that \( l^2 + m^2 + n^2 = 1 \) always holds for direction cosines - this is the key constraint when one or more angles are unknown.

Question 17. Find the angle between \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \), if \( \vec{a} = (2\hat{i} - \hat{j} + 3\hat{k}) \) and \( \vec{b} = (3\hat{i} + \hat{j} + 2\hat{k}) \).
Answer: We have \( \vec{a} = (2\hat{i} - \hat{j} + 3\hat{k}) \) and \( \vec{b} = (3\hat{i} + \hat{j} + 2\hat{k}) \). Calculate: \( \vec{a} + \vec{b} = (2\hat{i} - \hat{j} + 3\hat{k}) + (3\hat{i} + \hat{j} + 2\hat{k}) = 5\hat{i} + 5\hat{k} \). Also: \( \vec{a} - \vec{b} = (2\hat{i} - \hat{j} + 3\hat{k}) - (3\hat{i} + \hat{j} + 2\hat{k}) = -\hat{i} - 2\hat{j} + \hat{k} \). Find the magnitudes: \( |\vec{a} + \vec{b}| = \sqrt{5^2 + 5^2} = \sqrt{50} \), \( |\vec{a} - \vec{b}| = \sqrt{(-1)^2 + (-2)^2 + 1^2} = \sqrt{6} \). Compute the dot product: \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = (5)(-1) + (0)(-2) + (5)(1) = -5 + 5 = 0 \). Using the angle formula: \( \cos\theta = \frac{0}{\sqrt{50}\sqrt{6}} = 0 \), so \( \theta = \cos^{-1}(0) = \frac{\pi}{2} \).
In simple words: Add the two vectors and subtract them to get two new vectors. Find the magnitude of each new vector and their dot product. When the dot product is zero, the vectors are perpendicular, and the angle between them is 90 degrees.

Exam Tip: A zero dot product immediately tells you the vectors are perpendicular - no need to compute the inverse cosine when you see a dot product of zero.