CBSE Class 12 Mathematics Vector Algebra VBQs Set 01

Questions

Question. Find the scalar components of the vector \(\vec{AB}\) with initial point \(A(2, 1)\) and terminal point \(B(-5, 7)\). 

Answer: Let \(\vec{AB} = (-5 - 2)\hat{i} + (7 - 1)\hat{j} = -7\hat{i} + 6\hat{j}\)
Hence, scalar components are \(-7, 6\).
[Note: If \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) then \(x, y, z\) are called scalar components and \(x\hat{i}, y\hat{j}, z\hat{k}\) are called vector components.]

Question. Find a vector in the direction of vector \(5\hat{i} - \hat{j} + 2\hat{k}\) which has a magnitude of 8 units.

Answer: \(|\vec{a}| = \sqrt{(5)^2 + (-1)^2 + (2)^2} = \sqrt{25 + 1 + 4} = \sqrt{30}\)
The unit vector in the direction of vector \(\vec{a}\) is
\(\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k})\)
Now, vector in the direction of \(\vec{a}\) having magnitude 8 units is
\(8\hat{a} = \frac{8}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k}) = \frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}\)

Question. Find the position vector of a point \(R\) which divides the line joining two points \(P\) and \(Q\) whose position vectors are \(\hat{i} + 2\hat{j} - \hat{k}\) and \(-\hat{i} + \hat{j} + \hat{k}\) respectively, in the ratio 2 : 1
(i) internally (ii) externally

Answer: (i) Let \(R\) be the point which divides the line joining the point \(P\) and \(Q\) internally in the ratio 2 : 1.
The position vector of point \(R = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2 + 1} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}\).
(ii) Let \(R\) be the point which divides the line joining the points \(P\) and \(Q\) externally in the ratio 2 : 1.
\(R = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2 - 1} = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{1} = -3\hat{i} + 3\hat{k}\)

Question. Find \(|\vec{x}|\), if for a unit vector \(\vec{a}, (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12\).

Answer: Here \(|\vec{a}| = 1\) and \((\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12\)
Now, \((\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12 \Rightarrow \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12\)
\(\Rightarrow |\vec{x}|^2 + \vec{x} \cdot \vec{a} - \vec{x} \cdot \vec{a} - |\vec{a}|^2 = 12\)
\(\Rightarrow |\vec{x}|^2 - |\vec{a}|^2 = 12 \Rightarrow |\vec{x}|^2 - (1)^2 = 12\)
\(\Rightarrow |\vec{x}|^2 = 12 + 1 \Rightarrow |\vec{x}|^2 = 13 \Rightarrow |\vec{x}| = \sqrt{13}\)

Question. Using vectors, prove that the points (2, -1, 3), (3, -5, 1) and (-1, 11, 9) are collinear. 

Answer: Let \(A (2, -1, 3)\), \(B (3, -5, 1)\) and \(C (-1, 11, 9)\) are three points.
To show that \(A, B, C\) are collinear.
∴ \(\vec{AB} = (3 - 2)\hat{i} + (-5 + 1)\hat{j} + (1 - 3)\hat{k} = \hat{i} - 4\hat{j} - 2\hat{k} = \sqrt{1^2 + (-4)^2 + (-2)^2} = \sqrt{21}\)
\(\vec{BC} = (-1 - 3)\hat{i} + (11 + 5)\hat{j} + (9 - 1)\hat{k} = -4\hat{i} + 16\hat{j} + 8\hat{k} = \sqrt{(-4)^2 + (16)^2 + (8)^2} = 4\sqrt{21}\)
\(\vec{AC} = (-1 - 2)\hat{i} + (11 + 1)\hat{j} + (9 - 3)\hat{k} = -3\hat{i} + 12\hat{j} + 6\hat{k} = \sqrt{(-3)^2 + (12)^2 + (6)^2} = 3\sqrt{21}\)
∵ \(|\vec{AC}| + |\vec{AB}| = |\vec{BC}|\)
\(\Rightarrow A, B, C\) are collinear.

Question. Find \(\lambda\) and \(\mu\) if \((2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}\).

Answer: Here \((2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}\)
∴ \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = \vec{0}\)
\(\Rightarrow (6\mu - 27\lambda)\hat{i} - (2\mu - 27)\hat{j} + (2\lambda - 6)\hat{k} = \vec{0}\)
\(\Rightarrow 6\mu - 27\lambda = 0, \quad 2\mu - 27 = 0 \text{ and } 2\lambda - 6 = 0\)
\(\Rightarrow \mu = \frac{27}{2} \text{ and } \lambda = 3\)

Question. Find a unit vector perpendicular to each of the vectors \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\), 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, \(\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\)
∴ \(\vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} + 0\hat{k}\) and \(\vec{a} - \vec{b} = 2\hat{i} + 0\hat{j} + 4\hat{k}\)
Now, vector perpendicular to \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\) is
\((\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = (16 - 0)\hat{i} - (16 - 0)\hat{j} + (0 - 8)\hat{k} = 16\hat{i} - 16\hat{j} - 8\hat{k}\)
∴ Unit vector perpendicular to \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\) is given by
\(\pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{\sqrt{16^2 + (-16)^2 + (-8)^2}} = \pm \frac{8(2\hat{i} - 2\hat{j} - \hat{k})}{8\sqrt{2^2 + 2^2 + 1^2}}\)
\(= \pm \frac{2\hat{i} - 2\hat{j} - \hat{k}}{\sqrt{9}} = \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) = \pm \frac{2}{3}\hat{i} \mp \frac{2}{3}\hat{j} \mp \frac{1}{3}\hat{k}\)

Question. The two adjacent sides of a parallelogram are \(2\hat{i} - 4\hat{j} + 5\hat{k}\) and \(\hat{i} - 2\hat{j} - 3\hat{k}\). Find the unit vector parallel to one of its diagonals. Also, find its area. 

Answer: Let two adjacent sides \(\vec{AB}\) and \(\vec{AC}\) of a parallelogram \(ABDC\) be represented by \(2\hat{i} - 4\hat{j} + 5\hat{k}\) and \(\hat{i} - 2\hat{j} - 3\hat{k}\) in magnitude and direction respectively.
i.e., \(\vec{AB} = 2\hat{i} - 4\hat{j} + 5\hat{k}\) and \(\vec{AC} = \hat{i} - 2\hat{j} - 3\hat{k}\)
By parallelogram law of addition
\(\vec{AD} = \vec{AB} + \vec{BD} \Rightarrow \vec{AD} = \vec{AB} + \vec{AC}\) [\(\because \vec{BD} = \vec{AC}\)]
\(\vec{AD} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (\hat{i} - 2\hat{j} - 3\hat{k}) = 3\hat{i} - 6\hat{j} + 2\hat{k}\)
\(|\vec{AD}| = |3\hat{i} - 6\hat{j} + 2\hat{k}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\)
Therefore, unit vector parallel to diagonal \(\vec{AD} = \frac{1}{|\vec{AD}|}\vec{AD} = \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) = \frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}\)
Again, \(\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} = (12 + 10)\hat{i} - (-6 - 5)\hat{j} + (-4 + 4)\hat{k} = 22\hat{i} + 11\hat{j}\)
\(|\vec{AB} \times \vec{AC}| = |22\hat{i} + 11\hat{j}| = \sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605} = 11\sqrt{5}\)
Now the area of parallelogram \(ABDC\) whose adjacent sides are \(\vec{AB}\) and \(\vec{AC}\) is \(= |\vec{AB} \times \vec{AC}| = 11\sqrt{5}\) sq units.

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 \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{c} \cdot \vec{d} = 15\). 

Answer: The vector \(\vec{d}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), so we have \(\vec{d} = \lambda(\vec{a} \times \vec{b})\).
Now \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = (28 + 4)\hat{i} - (7 - 6)\hat{j} + (-2 - 12)\hat{k} = 32\hat{i} - \hat{j} - 14\hat{k}\)
So, \(\vec{d} = 32\lambda\hat{i} - \lambda\hat{j} - 14\lambda\hat{k}\)
We have \(\vec{c} \cdot \vec{d} = 15 \Rightarrow (2\hat{i} - \hat{j} + 4\hat{k}) \cdot (32\lambda\hat{i} - \lambda\hat{j} - 14\lambda\hat{k}) = 15\)
\(\Rightarrow 64\lambda + \lambda - 56\lambda = 15 \Rightarrow 9\lambda = 15 \Rightarrow \lambda = \frac{15}{9} = \frac{5}{3}\)
∴ Required vector \(\vec{d} = \frac{5}{3}(32\hat{i} - \hat{j} - 14\hat{k}) = \frac{160}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{70}{3}\hat{k}\).

Question. The scalar product of the vector \(\hat{i} + \hat{j} + \hat{k}\) with the 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 one. Find the value of \(\lambda\). 

Answer: Let sum of vectors \(2\hat{i} + 4\hat{j} - 5\hat{k}\) and \(\lambda\hat{i} + 2\hat{j} + 3\hat{k}\) equal to \(\vec{a}\) then \(\vec{a} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}\)
The unit vector in the direction of \(\vec{a} = \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 36 + 4}}\)
Here, \((\hat{i} + \hat{j} + \hat{k}) \cdot \hat{a} = 1 \Rightarrow (\hat{i} + \hat{j} + \hat{k}) \cdot \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 40}} = 1\)
\(\Rightarrow \frac{(2 + \lambda) + 6 - 2}{\sqrt{(2 + \lambda)^2 + 40}} = 1 \Rightarrow (\lambda + 6)^2 = (2 + \lambda)^2 + 40\)
\(\Rightarrow \lambda^2 + 36 + 12\lambda = 4 + \lambda^2 + 4\lambda + 40 \Rightarrow 8\lambda = 8 \Rightarrow \lambda = 1\).

Question. If with reference to the right handed system of mutually perpendicular unit vectors \(\hat{i}, \hat{j}\) and \(\hat{k}\), \(\vec{\alpha} = 3\hat{i} - \hat{j}\), \(\vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k}\), then express \(\vec{\beta}\) in the form \(\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2\), where \(\vec{\beta}_1\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\).

Answer: Let \(\vec{\beta}_1 = \lambda\vec{\alpha}\), \(\lambda\) is a scalar, i.e., \(\vec{\beta}_1 = 3\lambda\hat{i} - \lambda\hat{j}\)
and \(\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = (2 - 3\lambda)\hat{i} + (1 + \lambda)\hat{j} - 3\hat{k}\)
Now, since \(\vec{\beta}_2\) is to be perpendicular to \(\vec{\alpha}\), we should have \(\vec{\alpha} \cdot \vec{\beta}_2 = 0\) i.e.,
\(3(2 - 3\lambda) - (1 + \lambda) = 0 \Rightarrow 6 - 9\lambda - 1 - \lambda = 0 \Rightarrow 10\lambda = 5 \Rightarrow \lambda = \frac{1}{2}\)
Therefore, \(\vec{\beta}_1 = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}\) and \(\vec{\beta}_2 = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}\)

Multiple Choice Questions 

Question. Choose and write the correct option in the following questions.

Question. The vectors \(3\hat{i} - \hat{j} + 2\hat{k}, 2\hat{i} + \hat{j} + 3\hat{k}\) and \(\hat{i} + \lambda\hat{j} - \hat{k}\) are coplanar if 
(a) -2
(b) 0
(c) 2
(d) Any real number
Answer: (a)

Question. The area of a triangle formed by vertices \(O, A, B\) where \(\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}\) and \(\vec{OB} = -3\hat{i} - 2\hat{j} + \hat{k}\) is 
(a) \(3\sqrt{5}\) sq. units
(b) \(5\sqrt{5}\) sq. units
(c) \(6\sqrt{5}\) sq. units
(d) 4 sq. units
Answer: (a)

Question. The value of \(\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})\) is
(a) 0
(b) -1
(c) 1
(d) 3
Answer: (c)

Question. If \(\theta\) is the angle between any two vectors \(\vec{a}\) and \(\vec{b}\) then \(|\vec{a} - \vec{b}| = |\vec{a} + \vec{b}|\), where \(\theta\) is equal to
(a) 0
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{2}\)
(d) \(\pi\)
Answer: (c)

Question. The vector of the direction of the vector \(\hat{i} - 2\hat{j} + 2\hat{k}\) that has magnitude 9 is 
(a) \(\hat{i} - 2\hat{j} + 2\hat{k}\)
(b) \(\frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}\)
(c) \(3(\hat{i} - 2\hat{j} + 2\hat{k})\)
(d) \(9(\hat{i} - 2\hat{j} + 2\hat{k})\)
Answer: (c)

Question. The position vector of the point which divides the join of point \( 2\vec{a}-3\vec{b} \) and \( \vec{a}+\vec{b} \) in the ratio 3 : 1 is 
(a) \( \frac{3\vec{a}-2\vec{b}}{2} \)
(b) \( \frac{7\vec{a}-8\vec{b}}{4} \)
(c) \( \frac{3\vec{a}}{4} \)
(d) \( \frac{5\vec{a}}{4} \)
Answer: (d)

Question. The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4) respectively is
(a) \( -\hat{i} + 12\hat{j} + 4\hat{k} \)
(b) \( 5\hat{i} + 2\hat{j} - 4\hat{k} \)
(c) \( -5\hat{i} + 2\hat{j} + 4\hat{k} \)
(d) \( \hat{i} + \hat{j} + \hat{k} \)
Answer: (c)

Question. The angle between two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes \( \sqrt{3} \) and 4 respectively and \( \vec{a} \cdot \vec{b} = 2\sqrt{3} \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{5\pi}{2} \)
Answer: (b)

Question. Find the value of \( \lambda \) such that the vectors \( \vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \) are orthogonal
(a) 0
(b) 1
(c) \( \frac{3}{2} \)
(d) \( -\frac{5}{2} \)
Answer: (d)

Question. The value of \( \lambda \) for which the vectors \( 3\hat{i} - 6\hat{j} + \hat{k} \) and \( 2\hat{i} - 4\hat{j} + \lambda\hat{k} \) are parallel is
(a) \( \frac{2}{3} \)
(b) \( \frac{3}{2} \)
(c) \( \frac{5}{2} \)
(d) \( \frac{2}{5} \)
Answer: (a)

Question. The vector from origin to the points A and B are \( \vec{a} = 2\hat{i} - 3\hat{j} + 2\hat{k} \) and \( \vec{b} = 2\hat{i} + 3\hat{j} + \hat{k} \), respectively then the area of triangle OAB is [NCERT Exemplar]
(a) 340
(b) \( \sqrt{25} \)
(c) \( \sqrt{229} \)
(d) \( \frac{1}{2}\sqrt{229} \)
Answer: (d)

Question. For any vector \( \vec{a} \), the value of \( (\vec{a} \times \hat{i})^2 + (\vec{a} \times \hat{j})^2 + (\vec{a} \times \hat{k})^2 \) is equal to 
(a) \( \vec{a}^2 \)
(b) \( 3\vec{a}^2 \)
(c) \( 4\vec{a}^2 \)
(d) \( 2\vec{a}^2 \)
Answer: (d)

Question. If \( |\vec{a}| = 10, |\vec{b}| = 2 \) and \( \vec{a} \cdot \vec{b} = 12 \), then value of \( |\vec{a} \times \vec{b}| \) is
(a) 5
(b) 10
(c) 14
(d) 16
Answer: (d)

Question. The vector \( \lambda\hat{i} + \hat{j} + 2\hat{k} \), \( \hat{i} + \lambda\hat{j} - \hat{k} \) and \( 2\hat{i} - \hat{j} + \lambda\hat{k} \) are coplanar if
(a) \( \lambda = -2 \)
(b) \( \lambda = 0 \)
(c) \( \lambda = 1 \)
(d) \( \lambda = -1 \)
Answer: (a)

Question. If \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors such that \( \vec{a} + \vec{b} + \vec{c} = 0 \), then the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) is
(a) 1
(b) 3
(c) \( -\frac{3}{2} \)
(d) None of these
Answer: (c)

Question. Projection vector of \( \vec{a} \) on \( \vec{b} \) is 
(a) \( \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b} \)
(b) \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \)
(c) \( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \)
(d) \( \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \hat{b} \)
Answer: (a)

Question. If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are three vectors such that \( \vec{a} + \vec{b} + \vec{c} = 0 \) and \( |\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5 \), then value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) is
(a) 0
(b) 1
(c) – 19
(d) 38
Answer: (c)

Question. If \( |\vec{a}| = 4 \) and \( -3 \le \lambda \le 2 \), then the range of \( |\lambda \vec{a}| \) is 
(a) [0, 8]
(b) [– 12, 8]
(c) [0, 12]
(d) [8, 12]
Answer: (c)

Question. The number of vectors of unit length perpendicular to the vectors \( \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{j} + \hat{k} \) is
(a) one
(b) two
(c) three
(d) infinite
Answer: (b)

Question. The position vector of the point which divides the join of points with position vectors \( \vec{a} + \vec{b} \) and \( 2\vec{a} - \vec{b} \) in the ratio 1 : 2 is 
(a) \( \frac{3\vec{a} + 2\vec{b}}{3} \)
(b) \( \vec{a} \)
(c) \( \frac{5\vec{a} - \vec{b}}{3} \)
(d) \( \frac{4\vec{a} + \vec{b}}{3} \)
Answer: (d)

Fill in the Blanks 

Question. The projection of the vector \( \hat{i} - \hat{j} \) on the vector \( \hat{i} + \hat{j} \) is _____________ . 
Answer: 0
Solution: Let \( \vec{a} = \hat{i} - \hat{j} \) and \( \vec{b} = \hat{i} + \hat{j} \)
\( \therefore \) Projection of \( \vec{a} \) on \( \vec{b} = \vec{a} \cdot \frac{\vec{b}}{|\vec{b}|} = (\hat{i} - \hat{j}) \cdot \frac{(\hat{i} + \hat{j})}{\sqrt{(1)^2 + (1)^2}} = \frac{1 - 1}{\sqrt{2}} = \frac{0}{\sqrt{2}} = 0 \)

Question. If \( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144 \) and \( |\vec{a}| = 4 \), then \( |\vec{b}| \) is equal to _____________ .
Answer: 3
Solution: We have, \( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144 \)
\( \Rightarrow (|\vec{a}| |\vec{b}| \sin \theta)^2 + (|\vec{a}| |\vec{b}| \cos \theta)^2 = 144 \)
\( \Rightarrow (4 |\vec{b}| \sin \theta)^2 + (4 |\vec{b}| \cos \theta)^2 = 144 \)
\( \Rightarrow 16 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta) = 144 \)
\( \Rightarrow 16 |\vec{b}|^2 = 144 \Rightarrow |\vec{b}|^2 = 9 \Rightarrow |\vec{b}| = 3 \)

Question. If \( \vec{a} \) is a non-zero vector, then \( (\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k} \) equals _____________ . 
Answer: \( \vec{a} \)

Question. If \( |\vec{a}| = 1, \text{ and } \vec{a} \times \hat{i} = \hat{j} \), then angle between \( \vec{a} \) and \( \hat{i} \) is _____________ .
Answer: \( \frac{\pi}{2} \)

Question. The area of the triangle whose adjacent sides are \( \vec{a} = \hat{i} + 4\hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + 2\hat{k} \) is _____________ sq. units.
Answer: \( \frac{3}{2}\sqrt{11} \) Sq. units.
Solution: We have, \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -1 \\ 1 & 1 & 2 \end{vmatrix} = 9\hat{i} - 3\hat{j} - 3\hat{k} \)
\( \therefore |\vec{a} \times \vec{b}| = \sqrt{(9)^2 + (-3)^2 + (-3)^2} = \sqrt{81 + 9 + 9} = \sqrt{99} = 3\sqrt{11} \)
\( \therefore \) Area of triangle \( = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{3}{2}\sqrt{11} \) Sq. units.