CBSE Class 12 Mathematics Vector Algebra Worksheet Set 02

Long Answer Questions-I 

Question. Prove that, for any three vectors \( \vec{a}, \vec{b}, \vec{c} \)
\( [\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}] = 2[\vec{a} \quad \vec{b} \quad \vec{c}] \) 
Answer: \( \text{LHS} = [\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}] = (\vec{a}+\vec{b}) \cdot \{(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})\} \)
\( = (\vec{a}+\vec{b}) \cdot \{\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}\} = (\vec{a}+\vec{b}) \cdot \{\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}\} \quad [\because \vec{c} \times \vec{c} = \vec{0}] \)
\( = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a}) \)
\( = [\vec{a} \quad \vec{b} \quad \vec{c}] + [\vec{a} \quad \vec{b} \quad \vec{a}] + [\vec{a} \quad \vec{c} \quad \vec{a}] + [\vec{b} \quad \vec{b} \quad \vec{c}] + [\vec{b} \quad \vec{b} \quad \vec{a}] + [\vec{b} \quad \vec{c} \quad \vec{a}] \)
\( = [\vec{a} \quad \vec{b} \quad \vec{c}] + 0 + 0 + 0 + 0 + [\vec{b} \quad \vec{c} \quad \vec{a}] \quad [\text{By property of scalar triple product}] \)
\( = [\vec{a} \quad \vec{b} \quad \vec{c}] + [\vec{b} \quad \vec{c} \quad \vec{a}] = [\vec{a} \quad \vec{b} \quad \vec{c}] + [\vec{a} \quad \vec{b} \quad \vec{c}] \quad [\text{By property of circularly rotation}] \)
\( = 2[\vec{a} \quad \vec{b} \quad \vec{c}] = \text{RHS} \)

Question. Find the value of x such that the point A(3, 2, 1), B(4, x, 5), C(4, 2, -2) and D(6, 5, -1) are coplanar. 
Answer: We have A(3, 2, 1), B(4, x, 5), C(4, 2, -2) and D(6, 5, -1) points.
\( \vec{AB} = \hat{i} + (x-2)\hat{j} + 4\hat{k}; \quad \vec{AC} = \hat{i} + 0\hat{j} - 3\hat{k}; \quad \vec{AD} = 3\hat{i} + 3\hat{j} - 2\hat{k} \)
\( \because \text{Points A, B, C and D are coplanar} \)
\( \implies \) \( \vec{AB}, \vec{AC}, \vec{AD} \text{ are coplanar} \)
\( \implies \) \( [\vec{AB} \quad \vec{AC} \quad \vec{AD}] = 0 \)
\( \implies \) \( \begin{vmatrix} 1 & x-2 & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \end{vmatrix} = 0 \)
\( \implies \) \( 1(0 + 9) - (x - 2)(-2 + 9) + 4(3 - 0) = 0 \)
\( \implies \) \( 9 - 7x + 14 + 12 = 0 \)
\( \implies \) \( 7x = 35 \)
\( \implies \) \( x = 5 \)

Question. Show that the vectors \( \vec{a}, \vec{b}, \vec{c} \) are coplanar, iff \( \vec{a}+\vec{b}, \vec{b}+\vec{c} \) and \( \vec{c}+\vec{a} \) are coplanar. 
Answer: If part: Let \( \vec{a}, \vec{b}, \vec{c} \) are coplanar
\( \implies \) \( \text{Scalar triple product of } \vec{a}, \vec{b} \text{ and } \vec{c} \text{ is zero.} \)
\( \implies \) \( [\vec{a} \quad \vec{b} \quad \vec{c}] = 0 \)
\( \implies \) \( \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b}) = 0 \)
Now, \( [\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}] = (\vec{a}+\vec{b}) \cdot \{(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})\} \)
\( = (\vec{a}+\vec{b}) \cdot \{\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}\} = (\vec{a}+\vec{b}) \cdot \{\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}\} \quad [\because \vec{c} \times \vec{c} = 0] \)
\( = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a}) \)
\( = [\vec{a} \quad \vec{b} \quad \vec{c}] + 0 + 0 + 0 + 0 + [\vec{b} \quad \vec{c} \quad \vec{a}] \quad [\text{By property of scalar triple product}] \)
\( = [\vec{a} \quad \vec{b} \quad \vec{c}] + [\vec{a} \quad \vec{b} \quad \vec{c}] = 2[\vec{a} \quad \vec{b} \quad \vec{c}] = 2 \times 0 = 0 \quad [\because [\vec{a} \quad \vec{b} \quad \vec{c}] = 0] \)
Hence, \( \vec{a}+\vec{b}, \vec{b}+\vec{c} \text{ and } \vec{c}+\vec{a} \) are coplanar.
Only if part: Let \( \vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a} \) are coplanar.
\( \implies \) \( [\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}] = 0 \)
\( \implies \) \( (\vec{a}+\vec{b}) \cdot \{(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})\} = 0 \)
\( \implies \) \( (\vec{a}+\vec{b}) \cdot \{\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}\} = 0 \)
\( \implies \) \( (\vec{a}+\vec{b}) \cdot \{\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}\} = 0 \quad [\because \vec{c} \times \vec{c} = \vec{0}] \)
\( \implies \) \( \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a}) = 0 \)
\( \implies \) \( [\vec{a} \quad \vec{b} \quad \vec{c}] + 0 + 0 + 0 + 0 + [\vec{b} \quad \vec{c} \quad \vec{a}] = 0 \)
\( \implies \) \( 2[\vec{a} \quad \vec{b} \quad \vec{c}] = 0 \quad [\because [\vec{a} \quad \vec{b} \quad \vec{c}] = [\vec{b} \quad \vec{c} \quad \vec{a}]] \)
\( \implies \) \( [\vec{a} \quad \vec{b} \quad \vec{c}] = 0 \)
Hence, \( \vec{a}, \vec{b}, \vec{c} \) are coplanar.

Question. Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} \) and \( \vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \) then
(a) Let \( c_1 = 1 \) and \( c_2 = 2 \), find \( c_3 \) which makes \( \vec{a}, \vec{b} \) and \( \vec{c} \) coplanar.
(b) If \( c_2 = -1 \) and \( c_3 = 1 \), show that no value of \( c_1 \) can make \( \vec{a}, \vec{b} \) and \( \vec{c} \) coplanar.
Answer: Given \( \vec{a} = \hat{i} + \hat{j} + \hat{k}; \quad \vec{b} = \hat{i} \text{ and } \vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \)
(a) Since \( \vec{a}, \vec{b} \text{ and } \vec{c} \) vectors are coplanar
\( \implies \) \( [\vec{a} \quad \vec{b} \quad \vec{c}] = 0 \)
\( \implies \) \( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0 \)
\( \implies \) \( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & c_3 \end{vmatrix} = 0 \quad [\text{Given that } c_1 = 1 \text{ and } c_2 = 2] \)
\( \implies \) \( 1(0 - 0) - 1(c_3 - 0) + 1(2 - 0) = 0 \)
\( \implies \) \( -c_3 + 2 = 0 \)
\( \implies \) \( c_3 = 2 \)
(b) To make \( \vec{a}, \vec{b} \text{ and } \vec{c} \) coplanar.
\( \implies \) \( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0 \)
\( \implies \) \( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ c_1 & -1 & 1 \end{vmatrix} = 0 \quad [\text{Given that } c_2 = -1 \text{ and } c_3 = 1] \)
\( \implies \) \( 1(0 - 0) - 1(1 - 0) + 1(-1 - 0) = 0 \)
\( \implies \) \( -1 - 1 = 0 \)
\( \implies \) \( -2 = 0, \text{ which is never possible.} \)
Hence, if \( c_2 = -1 \text{ and } c_3 = 1 \), there is no value of \( c_1 \) which can make \( \vec{a}, \vec{b} \text{ and } \vec{c} \) coplanar.

Question. If \( \vec{a}, \vec{b}, \vec{c} \) are mutually perpendicular vectors of equal magnitudes, show that the vector \( \vec{a} + \vec{b} + \vec{c} \) is equally inclined to \( \vec{a}, \vec{b} \) and \( \vec{c} \). Also, find the angle which \( \vec{a} + \vec{b} + \vec{c} \) makes with \( \vec{a} \) or \( \vec{b} \) or \( \vec{c} \). 
Answer: Let \( |\vec{a}| = |\vec{b}| = |\vec{c}| = x \text{ (say)} \)
Since \( \vec{a}, \vec{b}, \vec{c} \) are mutually perpendicular vectors.
Therefore, \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0 = \vec{b} \cdot \vec{a} = \vec{c} \cdot \vec{b} = \vec{a} \cdot \vec{c} \)
Now, \( |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \)
\( = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} \)
\( = x^2 + 0 + 0 + 0 + x^2 + 0 + 0 + 0 + x^2 = 3x^2 \)
\( \implies \) \( |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}x \)
Let \( \theta_1, \theta_2 \text{ and } \theta_3 \) be the angles made by \( (\vec{a} + \vec{b} + \vec{c}) \) with \( \vec{a}, \vec{b} \text{ and } \vec{c} \) respectively.
\( \therefore \cos \theta_1 = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{a}| \cdot |\vec{a} + \vec{b} + \vec{c}|} = \frac{\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}}{x \cdot \sqrt{3}x} = \frac{x^2 + 0 + 0}{\sqrt{3}x^2} = \frac{1}{\sqrt{3}} \)
\( \implies \) \( \theta_1 = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \text{ similarly } \theta_2 = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \text{ and } \theta_3 = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \)
i.e., \( (\vec{a} + \vec{b} + \vec{c}) \) is equally inclined with \( \vec{a}, \vec{b} \text{ and } \vec{c} \)
\( \implies \) \( 42 + 14\lambda = 0 \)
\( \implies \) \( 14\lambda = -42 \)
\( \implies \) \( \lambda = -3 \)

Question. If \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{j} - \hat{k} \), then find a vector \( \vec{c} \) such that \( \vec{a} \times \vec{c} = \vec{b} \) and \( \vec{a} \cdot \vec{c} = 3 \).
Answer: Let \( \vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}. \) Then,
\( (\vec{a} \times \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ c_1 & c_2 & c_3 \end{vmatrix} = (c_3 - c_2)\hat{i} + (c_1 - c_3)\hat{j} + (c_2 - c_1)\hat{k} \)
\( \because (\vec{a} \times \vec{c}) = \vec{b} \)
\( \implies \) \( (c_3 - c_2)\hat{i} + (c_1 - c_3)\hat{j} + (c_2 - c_1)\hat{k} = \hat{j} - \hat{k} \)
\( \implies \) \( c_3 - c_2 = 0, \quad c_1 - c_3 = 1 \quad \text{and} \quad c_2 - c_1 = -1 \quad ... (i) \)
Also, \( \vec{a} \cdot \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \cdot (c_1\hat{i} + c_2\hat{j} + c_3\hat{k}) \)
\( \implies \) \( \vec{a} \cdot \vec{c} = c_1 + c_2 + c_3 \)
\( \implies \) \( c_1 + c_2 + c_3 = 3 \quad [\because \vec{a} \cdot \vec{c} = 3] \quad ... (ii) \)
\( \implies \) \( c_1 + c_2 + c_1 - 1 = 3 \quad [\because c_1 - c_3 = 1] \quad ... (iii) \)
\( \implies \) \( 2c_1 + c_2 = 4 \)
On solving \( c_1 - c_2 = 1 \) and \( 2c_1 + c_2 = 4 \), we get
\( 3c_1 = 5 \)
\( \implies \) \( c_1 = \frac{5}{3} \)
\( \therefore c_2 = (c_1 - 1) = \left(\frac{5}{3} - 1\right) = \frac{2}{3} \quad \text{and} \quad c_3 = c_2 = \frac{2}{3} \)
Hence, \( \vec{c} = \left(\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\right). \)

Question. If \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) and \( |\vec{a}| = 3, |\vec{b}| = 5 \) and \( |\vec{c}| = 7 \) then show that the angle between \( \vec{a} \) and \( \vec{b} \) is \( 60^\circ \). 
Answer: \( \vec{a} + \vec{b} + \vec{c} = 0 \)
\( \implies \) \( (\vec{a} + \vec{b})^2 = (-\vec{c})^2 \)
\( \implies \) \( (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{c} \cdot \vec{c} \)
\( \implies \) \( |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{c}|^2 \)
\( \implies \) \( 9 + 25 + 2\vec{a} \cdot \vec{b} = 49 \)
\( \implies \) \( 2\vec{a} \cdot \vec{b} = 49 - 25 - 9 \)
\( \implies \) \( 2|\vec{a}||\vec{b}| \cos \theta = 15 \)
\( \implies \) \( 30 \cos \theta = 15 \)
\( \implies \) \( \cos \theta = \frac{1}{2} = \cos 60^\circ \)
\( \implies \) \( \theta = 60^\circ \)

Question. If \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), then prove that \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \), and hence show that \( [\vec{a} \quad \vec{b} \quad \vec{c}] = 0 \).
Answer: Given \( \vec{a} + \vec{b} + \vec{c} = 0 \)
\( \implies \) \( \vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times \vec{0} \)
\( \implies \) \( \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0} \)
\( \implies \) \( \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \quad ... (i) \)
Again, \( \vec{b} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{b} \times \vec{0} \)
\( \implies \) \( \vec{b} \times \vec{a} + \vec{b} \times \vec{c} = \vec{0} \)
\( \implies \) \( \vec{b} \times \vec{c} = \vec{a} \times \vec{b} \quad ... (ii) \)
From (i) and (ii), we get
\( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \)
Now, \( [\vec{a} \quad \vec{b} \quad \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) \)
\( = \vec{a} \cdot (\vec{a} \times \vec{b}) = [\vec{a} \quad \vec{a} \quad \vec{b}] = 0 \quad [\because \text{Scalar triple product of three vectors is zero if any two of them are equal.}] \)

Question. Let \( \vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}, \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k} \) and \( \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \). Find a vector \( \vec{p} \) which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and \( \vec{p} \cdot \vec{c} = 18 \).
Answer: Given, \( \vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}, \quad \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}, \quad \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \)
Vector \( \vec{p} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \) i.e., \( \vec{p} \) is parallel to vector \( \vec{a} \times \vec{b} \).
\( \therefore \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = \hat{i} \begin{vmatrix} 4 & 2 \\ -2 & 7 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 3 & 7 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 4 \\ 3 & -2 \end{vmatrix} = 32\hat{i} - \hat{j} - 14\hat{k} \)
Since \( \vec{p} \) is parallel to \( \vec{a} \times \vec{b} \)
\( \implies \) \( \vec{p} = \mu(32\hat{i} - \hat{j} - 14\hat{k}) \)
Also, \( \vec{p} \cdot \vec{c} = 18 \)
\( \implies \) \( \mu(32\hat{i} - \hat{j} - 14\hat{k}) \cdot (2\hat{i} - \hat{j} + 4\hat{k}) = 18 \)
\( \implies \) \( \mu(64 + 1 - 56) = 18 \)
\( \implies \) \( 9\mu = 18 \text{ or } \mu = 2 \)
\( \therefore \vec{p} = 2(32\hat{i} - \hat{j} - 14\hat{k}) = 64\hat{i} - 2\hat{j} - 28\hat{k} \)

Question. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \).
Answer: Given, two vectors are \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \text{ and } \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \)
If \( \vec{c} \) is the resultant vector of \( \vec{a} \) and \( \vec{b} \) then
\( \vec{c} = \vec{a} + \vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) + (\hat{i} - 2\hat{j} + \hat{k}) = 3\hat{i} + \hat{j} + 0\cdot\hat{k} \)
Now, a vector having magnitude 5 and parallel to \( \vec{c} \) is given by
\( \frac{5\vec{c}}{|\vec{c}|} = \frac{5(3\hat{i} + \hat{j} + 0\hat{k})}{\sqrt{3^2 + 1^2 + 0^2}} = \frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j} \)
It is required vector.
[Note: A vector having magnitude \( l \) and parallel to \( \vec{a} \) is given by \( l \cdot \frac{\vec{a}}{|\vec{a}|} \).]

Question. If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a} + \vec{b}| = |\vec{a}| \), then prove that vector \( 2\vec{a} + \vec{b} \) is perpendicular to vector \( \vec{b} \). 
Answer: \( \because |\vec{a} + \vec{b}| = |\vec{a}| \)
\( \implies \) \( |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 \)
\( \implies \) \( (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 \)
\( \implies \) \( \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = |\vec{a}|^2 \)
\( \implies \) \( |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = |\vec{a}|^2 \quad [\because \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}] \)
\( \implies \) \( 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 0 \)
\( \implies \) \( (2\vec{a} + \vec{b}) \cdot \vec{b} = 0 \)
\( \implies \) \( (2\vec{a} + \vec{b}) \text{ is perpendicular to } \vec{b}. \)

Question. The magnitude of the vector product of the vector \( \hat{i} + \hat{j} + \hat{k} \) with a unit vector along the sum of vectors \( 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \lambda\hat{i} + 2\hat{j} + 3\hat{k} \) is equal to \( \sqrt{2} \). Find the value of \( \lambda \). 
Answer: Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k}; \quad \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}; \quad \vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k} \)
From question
\( \left| \vec{a} \times \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} \right| = \sqrt{2} \)
\( \implies \) \( \frac{|\vec{a} \times (\vec{b} + \vec{c})|}{|\vec{b} + \vec{c}|} = \sqrt{2} \quad ... (i) \)
\( \vec{b} + \vec{c} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \)
\( \therefore |\vec{b} + \vec{c}| = \sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2} = \sqrt{4 + \lambda^2 + 4\lambda + 36 + 4} = \sqrt{\lambda^2 + 4\lambda + 44} \)
\( \implies \) \( \vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2+\lambda & 6 & -2 \end{vmatrix} = (-2-6)\hat{i} - (-2-2-\lambda)\hat{j} + (6-2-\lambda)\hat{k} = -8\hat{i} + (4+\lambda)\hat{j} + (4-\lambda)\hat{k} \)
Putting it in (i), we get
\( \frac{| -8\hat{i} + (4+\lambda)\hat{j} + (4-\lambda)\hat{k} |}{\sqrt{\lambda^2 + 4\lambda + 44}} = \sqrt{2} \)
\( \implies \) \( \frac{\sqrt{(-8)^2 + (4+\lambda)^2 + (4-\lambda)^2}}{\sqrt{\lambda^2 + 4\lambda + 44}} = \sqrt{2} \)
Squaring both sides, we get
\( \frac{64 + 16 + \lambda^2 + 8\lambda + 16 + \lambda^2 - 8\lambda}{\lambda^2 + 4\lambda + 44} = 2 \)
\( \implies \) \( \frac{96 + 2\lambda^2}{\lambda^2 + 4\lambda + 44} = 2 \)
\( \implies \) \( 8\lambda = 8 \)
\( \implies \) \( \lambda = 1 \)

Question. Show that the points A, B, C with position vectors \( 2\hat{i} - \hat{j} + \hat{k}, \hat{i} - 3\hat{j} - 5\hat{k} \) and \( 3\hat{i} - 4\hat{j} - 4\hat{k} \) respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle. 
Answer: Given, position vector of A \( = 2\hat{i} - \hat{j} + \hat{k} \)
position vector of B \( = \hat{i} - 3\hat{j} - 5\hat{k} \)
position vector of C \( = 3\hat{i} - 4\hat{j} - 4\hat{k} \)
\( \implies \) \( \vec{AB} = -\hat{i} - 2\hat{j} - 6\hat{k}; \quad \vec{AC} = \hat{i} - 3\hat{j} - 5\hat{k} \quad \text{and} \quad \vec{BC} = 2\hat{i} - \hat{j} + \hat{k} \)
Now, \( |\vec{AB}|^2 = \vec{AB} \cdot \vec{AB} = 1 + 4 + 36 = 41; \quad |\vec{AC}|^2 = 1 + 9 + 25 = 35; \quad |\vec{BC}|^2 = 4 + 1 + 1 = 6 \)
\( \because |\vec{AB}|^2 = |\vec{AC}|^2 + |\vec{BC}|^2 \)
\( \implies \) \( \text{A, B, C are the vertices of right triangle.} \)
Now, \( \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -2 & -6 \\ 1 & -3 & -5 \end{vmatrix} = \hat{i}(10 - 18) - \hat{j}(5 + 6) + \hat{k}(3 + 2) = -8\hat{i} - 11\hat{j} + 5\hat{k} \)
\( \therefore |\vec{AB} \times \vec{AC}| = \sqrt{(-8)^2 + (-11)^2 + 5^2} = \sqrt{64 + 121 + 25} = \sqrt{210} \)
\( \therefore \text{Area } (\Delta ABC) = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{\sqrt{210}}{2} \text{ sq. units} \)
Alternate method to find area:
\( \text{Area of } \Delta ABC = \frac{1}{2} \times |\vec{BC}| \times |\vec{AC}| = \frac{1}{2} \times \sqrt{35} \times \sqrt{6} = \frac{\sqrt{210}}{2} \text{ sq. units} \)

Question. Find a unit vector perpendicular to each of the vectors \( \vec{a} + 2\vec{b} \) and \( 2\vec{a} + \vec{b} \), where \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \).
Answer: Given, \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \text{ and } \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \)
\( \vec{a} + 2\vec{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) + (2\hat{i} + 4\hat{j} - 4\hat{k}) = 5\hat{i} + 6\hat{j} - 2\hat{k} \)
\( 2\vec{a} + \vec{b} = (6\hat{i} + 4\hat{j} + 4\hat{k}) + (\hat{i} + 2\hat{j} - 2\hat{k}) = 7\hat{i} + 6\hat{j} + 2\hat{k} \)
Now, perpendicular vector of \( (\vec{a} + 2\vec{b}) \text{ and } (2\vec{a} + \vec{b}) \)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 6 & -2 \\ 7 & 6 & 2 \end{vmatrix} = (12+12)\hat{i} - (10+14)\hat{j} + (30-42)\hat{k} = 24\hat{i} - 24\hat{j} - 12\hat{k} = 12(2\hat{i} - 2\hat{j} - \hat{k}) \)
\( \text{Required unit vector} = \pm \frac{12(2\hat{i} - 2\hat{j} - \hat{k})}{12\sqrt{2^2 + (-2)^2 + (-1)^2}} = \pm \frac{2\hat{i} - 2\hat{j} - \hat{k}}{3} = \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) \)

Question. If \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), find \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy \). 
Answer: Here, \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \)
Now, \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy = \{(x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{i}\} \cdot \{(x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{j}\} + xy \)
\( = (-y\hat{k} + z\hat{j}) \cdot (x\hat{k} - z\hat{i}) + xy = (0\hat{i} + z\hat{j} - y\hat{k}) \cdot (-z\hat{i} + 0\hat{j} + x\hat{k}) + xy \)
\( = 0 + 0 - xy + xy = 0 \)

Question. Find the area of a parallelogram ABCD whose side AB and the diagonal AC are given by the vectors \( 3\hat{i} + \hat{j} + 4\hat{k} \) and \( 4\hat{i} + 5\hat{k} \) respectively. 
Answer: Here, \( \vec{BC} = \vec{BA} + \vec{AC} = -\vec{AB} + \vec{AC} \)
\( = -3\hat{i} - \hat{j} - 4\hat{k} + 4\hat{i} + 5\hat{k} = \hat{i} - \hat{j} + \hat{k} \)
\( \therefore \vec{AD} = \vec{BC} = \hat{i} - \hat{j} + \hat{k} \)
\( \therefore \text{Area of parallelogram} = |\vec{AB} \times \vec{AD}| \)
\( = \left| \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} \right| = \left| (1+4)\hat{i} - (3-4)\hat{j} + (-3-1)\hat{k} \right| = |5\hat{i} + \hat{j} - 4\hat{k}| \)
\( = \sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42} \text{ sq. units.} \)

Question. If \( \vec{a} = 2\hat{i} - \hat{j} - 2\hat{k} \) and \( \vec{b} = 7\hat{i} + 2\hat{j} - 3\hat{k} \) then express \( \vec{b} \) in the from of \( \vec{b} = \vec{b}_1 + \vec{b}_2 \), where \( \vec{b}_1 \) is parallel to \( \vec{a} \) and \( \vec{b}_2 \) is perpendicular to \( \vec{a} \). 
Answer: Since \( \vec{b}_1 \parallel \vec{a} \)
\( \implies \) \( \vec{b}_1 = \lambda\vec{a} = \lambda(2\hat{i} - \hat{j} - 2\hat{k}) = 2\lambda\hat{i} - \lambda\hat{j} - 2\lambda\hat{k} \)
\( \because \vec{b}_1 + \vec{b}_2 = \vec{b} \)
\( \implies \) \( \vec{b}_2 = \vec{b} - \vec{b}_1 \)
\( = (7\hat{i} + 2\hat{j} - 3\hat{k}) - (2\lambda\hat{i} - \lambda\hat{j} - 2\lambda\hat{k}) = 7\hat{i} + 2\hat{j} - 3\hat{k} - 2\lambda\hat{i} + \lambda\hat{j} + 2\lambda\hat{k} \)
\( = (7 - 2\lambda)\hat{i} + (2 + \lambda)\hat{j} - (3 - 2\lambda)\hat{k} \)
It is given that \( \vec{b}_2 \) is perpendicular to \( \vec{a}. \)
\( \implies \) \( \vec{b}_2 \cdot \vec{a} = 0 \)
\( \implies \) \( (7 - 2\lambda) \cdot 2 - (2 + \lambda) \cdot 1 + (3 - 2\lambda) \cdot 2 = 0 \)
\( \implies \) \( 14 - 4\lambda - 2 - \lambda + 6 - 4\lambda = 0 \)
\( \implies \) \( -9\lambda + 18 = 0 \)
\( \implies \) \( \lambda = \frac{18}{9} = 2 \)
Hence, \( \vec{b}_1 = 4\hat{i} - 2\hat{j} - 4\hat{k}; \quad \vec{b}_2 = 3\hat{i} + 4\hat{j} + \hat{k} \)
Now, \( 7\hat{i} + 2\hat{j} - 3\hat{k} = (4\hat{i} - 2\hat{j} - 4\hat{k}) + (3\hat{i} + 4\hat{j} + \hat{k}), \text{ i.e., } \vec{b} = \vec{b}_1 + \vec{b}_2 \)

Question. If \( \vec{a} \) and \( \vec{b} \) are unit vectors, then what is the angle between \( \vec{a} \) and \( \vec{b} \) for \( \vec{a} - \sqrt{2}\vec{b} \) to be a unit vector?
Answer: Given, \( \vec{a} - \sqrt{2}\vec{b} \) is an unit vector
\( \implies \) \( |\vec{a} - \sqrt{2}\vec{b}| = 1 \)
\( \implies \) \( |\vec{a} - \sqrt{2}\vec{b}|^2 = 1 \)
\( \implies \) \( (\vec{a} - \sqrt{2}\vec{b}) \cdot (\vec{a} - \sqrt{2}\vec{b}) = 1 \)
\( \implies \) \( \vec{a} \cdot \vec{a} - \sqrt{2}\vec{a} \cdot \vec{b} - \sqrt{2}\vec{b} \cdot \vec{a} + 2\vec{b} \cdot \vec{b} = 1 \)
\( \implies \) \( |\vec{a}|^2 - 2\sqrt{2}\vec{a} \cdot \vec{b} + 2|\vec{b}|^2 = 1 \)
\( \implies \) \( [\because \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}] \)
\( \implies \) \( 1 - 2\sqrt{2}\vec{a} \cdot \vec{b} + 2 = 1 \)
\( \implies \) \( [\because |\vec{a}| = |\vec{b}| = 1] \)
\( \implies \) \( -2\sqrt{2}\vec{a} \cdot \vec{b} = -2 \)
\( \implies \) \( \vec{a} \cdot \vec{b} = \frac{-2}{-2\sqrt{2}} \)
\( \implies \) \( \vec{a} \cdot \vec{b} = \frac{1}{\sqrt{2}} \)
\( \implies \) \( |\vec{a}||\vec{b}| \cos \theta = \frac{1}{\sqrt{2}} \quad [\because \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta] \)
\( \implies \) \( 1 \cdot 1 \cdot \cos \theta = \frac{1}{\sqrt{2}} \)
\( \implies \) \( \cos \theta = \cos \frac{\pi}{4} \)
\( \implies \) \( \theta = \frac{\pi}{4} \)

Question. If \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \), then show that \( (\vec{a} - \vec{d}) \) is parallel to \( (\vec{b} - \vec{c}) \), it is being given that \( \vec{a} \neq \vec{d} \) and \( \vec{b} \neq \vec{c} \).
Answer: Given, \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \text{ and } \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \)
\( \implies \) \( \vec{a} \times \vec{b} - \vec{a} \times \vec{c} = \vec{c} \times \vec{d} - \vec{b} \times \vec{d} \)
\( \implies \) \( \vec{a} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{d} - \vec{c} \times \vec{d} = \vec{0} \)
\( \implies \) \( \vec{a} \times (\vec{b} - \vec{c}) + (\vec{b} - \vec{c}) \times \vec{d} = \vec{0} \quad [\text{By left and right distributive law}] \)
\( \implies \) \( \vec{a} \times (\vec{b} - \vec{c}) - \vec{d} \times (\vec{b} - \vec{c}) = \vec{0} \quad [\because \vec{a} \times \vec{b} = -\vec{b} \times \vec{a}] \)
\( \implies \) \( (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{0} \quad [\text{By right distributive law}] \)
\( \implies \) \( (\vec{a} - \vec{d}) \parallel (\vec{b} - \vec{c}) \)

Question. Prove that : \( |\vec{a} \times \vec{b}|^2 = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} \end{vmatrix} \)
Answer: Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \). Then,
\( \text{LHS} = |\vec{a} \times \vec{b}|^2 = (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) \)
\( = (ab \sin \theta) \hat{n} \cdot (ab \sin \theta) \hat{n} = (a^2b^2 \sin^2 \theta)(\hat{n} \cdot \hat{n}) = a^2b^2 \sin^2 \theta \)
\( = a^2b^2(1 - \cos^2 \theta) = a^2b^2 - (ab \cos \theta)^2 \)
\( = (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b})^2 \quad ... (i) \)
Also, \( \text{RHS} = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} \end{vmatrix} = (\vec{a} \cdot \vec{a}) \cdot (\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b}) \cdot (\vec{a} \cdot \vec{b}) \)
\( = (\vec{a} \cdot \vec{a}) \cdot (\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b})^2 \quad ... (ii) \)
From (i) and (ii) \( \text{RHS} = \text{LHS} \quad \text{Hence proved.} \)

Question. If \( \vec{a}, \vec{b} \) are unit vectors such that the vector \( \vec{a} + 3\vec{b} \) is perpendicular to \( 7\vec{a} - 5\vec{b} \) and \( \vec{a} - 4\vec{b} \) is perpendicular to \( 7\vec{a} - 2\vec{b} \), then find the angle between \( \vec{a} \) and \( \vec{b} \).
Answer: Let angle between \( \vec{a} \) and \( \vec{b} \) be \( \theta \)
Given, \( (\vec{a} + 3\vec{b}) \perp (7\vec{a} - 5\vec{b}) \)
\( \implies \) \( (\vec{a} + 3\vec{b}) \cdot (7\vec{a} - 5\vec{b}) = 0 \)
\( \implies \) \( 7|\vec{a}|^2 + 16(\vec{a} \cdot \vec{b}) - 15|\vec{b}|^2 = 0 \)
\( \implies \) \( 7 + 16 \cos \theta - 15 = 0 \quad [\because |\vec{a}|^2 = |\vec{b}|^2 = 1] \)
\( \implies \) \( \cos \theta = \frac{8}{16} = \frac{1}{2} \)
\( \implies \) \( \theta = \frac{\pi}{3} \)
Also, given that \( (\vec{a} - 4\vec{b}) \perp (7\vec{a} - 2\vec{b