CBSE Class 12 Mathematics HOTs Vector Algebra Set 02

Very Short Answer Type Questions

Question. If \(\vec{a}\) is a non-zero vector, then find the value of \((\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k}\). CBSE 2020
Answer: Let \(\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}\) \[\therefore (\vec{a} \cdot \hat{i}) = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot \hat{i} = x\] Similarly, \((\vec{a} \cdot \hat{j}) = y\) and \((\vec{a} \cdot \hat{k}) = z\)
Now, \((\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k} = x\hat{i} + y\hat{j} + z\hat{k} = \vec{a}\)

Question. Find the magnitude of projection of \((2\hat{i} + \hat{j} + \hat{k})\) on \((\hat{i} - 2\hat{j} + 2\hat{k})\). CBSE Sample Paper 2020
Answer: Projection of \((2\hat{i} + \hat{j} + \hat{k})\) on \((\hat{i} - 2\hat{j} + 2\hat{k})\) is \[\frac{(2\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + 2\hat{k})}{|\hat{i} - 2\hat{j} + 2\hat{k}|} = \frac{2 - 2 + 2}{\sqrt{1 + (-2)^2 + (2)^2}}\] \[= \frac{2}{\sqrt{9}} = \frac{2}{3}\] \(\therefore\) Required magnitude is \(\frac{2}{3}\) unit.

Question. Find \(|\vec{x}|\), if \((\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12\text{,}\) where \(\vec{a}\) is a unit vector. CBSE Sample Paper 2023
Answer: Given, \(\vec{a}\) is a unit vector. \[\therefore |\vec{a}| = 1\] Now, we have \[(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12\] \[\Rightarrow \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12\] \[\Rightarrow |\vec{x}|^2 - |\vec{a}|^2 = 12 \text{ [}\because \vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}\text{]}\] \[\Rightarrow |\vec{x}|^2 - 1 = 12 \text{ [}\because |\vec{a}| = 1 \text{ (given)]}\] \[\Rightarrow |\vec{x}|^2 = 13\] \[\Rightarrow |\vec{x}| = \sqrt{13}\]

Question. If \(\vec{a} + \vec{b} + \vec{c} = 0\) and \(|\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7\), then find the value of \(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}\). Competency Based Que
Answer: We have, \(\vec{a} + \vec{b} + \vec{c} = 0\) and \(|\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7\)
Now, consider \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0\) \[\Rightarrow \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 0\] \[\Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0\] \[\Rightarrow 3^2 + 5^2 + 7^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0\] \[\Rightarrow 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -(9 + 25 + 49)\] \[\Rightarrow (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -\frac{83}{2}\]

Question. If the sum of two unit vectors \(\hat{a}\) and \(\hat{b}\) is a unit vector, then show that the magnitude of their difference is \(\sqrt{3}\). CBSE 2019, 12C
Answer: Let \(\vec{c} = \hat{a} + \hat{b}\). Then, according to given condition \(\vec{c}\) is a unit vector i.e. \(|\vec{c}| = 1\).
To show \(|\hat{a} - \hat{b}| = \sqrt{3}\)
Consider, \(\vec{c} = \hat{a} + \hat{b} \Rightarrow |\vec{c}| = |\hat{a} + \hat{b}|\) \[\Rightarrow 1 = |\hat{a} + \hat{b}| \Rightarrow |\hat{a} + \hat{b}|^2 = 1\] \[\Rightarrow (\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b}) = 1\] \[\Rightarrow |\hat{a}|^2 + 2\hat{a} \cdot \hat{b} + |\hat{b}|^2 = 1\] \[\Rightarrow 1 + 2\hat{a} \cdot \hat{b} + 1 = 1\] \[\Rightarrow 2\hat{a} \cdot \hat{b} = -1 \text{ ...(i)}\] Now, consider \(|\hat{a} - \hat{b}|^2 = (\hat{a} - \hat{b}) \cdot (\hat{a} - \hat{b})\) \[= |\hat{a}|^2 - 2\hat{a} \cdot \hat{b} + |\hat{b}|^2\] \[= 1 - (-1) + 1 \text{ [using Eq. (i)]}\] \[\Rightarrow |\hat{a} - \hat{b}|^2 = 3\] \[\Rightarrow |\hat{a} - \hat{b}| = \sqrt{3} \text{ [taking positive square root, as magnitude cannot be negative]}\] Hence proved.

Question. If \(\vec{a} + \vec{b} + \vec{c} = 0\) and \(|\vec{a}| = 5, |\vec{b}| = 6\) and \(|\vec{c}| = 9\), then find the angle between \(\vec{a}\) and \(\vec{b}\). CBSE 2018C
Answer: Given, \(\vec{a} + \vec{b} + \vec{c} = 0\) \[\Rightarrow \vec{a} + \vec{b} = -\vec{c}\] \[\Rightarrow (\vec{a} + \vec{b})^2 = (-\vec{c})^2 \text{ [squaring both sides]}\] \[\Rightarrow (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c})\] \[\Rightarrow \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c}\] \[\Rightarrow |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 = |\vec{c}|^2 \text{ [}\because \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\text{]}\] \[\Rightarrow |\vec{a}|^2 + 2|\vec{a}||\vec{b}|\cos\theta + |\vec{b}|^2 = |\vec{c}|^2 \text{ ...(i) [}\because \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\text{]}\] On putting the values of \(|\vec{a}| = 5, |\vec{b}| = 6\) and \(|\vec{c}| = 9\) in Eq. (i), we get \[(5)^2 + 2 \times 5 \times 6 \times \cos\theta + (6)^2 = (9)^2\] \[25 + 60\cos\theta + 36 = 81\] \[60\cos\theta = 81 - 61 = 20\] \[\cos\theta = \frac{20}{60} = \frac{1}{3}\] \[\therefore \theta = \cos^{-1}\left(\frac{1}{3}\right)\]

Question. Find the projection (vector) of \(2\hat{i} - \hat{j} + \hat{k}\) on \(\hat{i} - 2\hat{j} + \hat{k}\).
Answer: Let \(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} - 2\hat{j} + \hat{k}\)
Now, projection vector of \(\vec{a}\) on \(\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}\)
Here, \(\vec{a} \cdot \vec{b} = (2\hat{i} - \hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k})\)
\(= 2 \times 1 + (-1) \times (-2) + 1 \times 1\)
\(= 2 + 2 + 1 = 5\)
and \(|\vec{b}| = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\)
\(\therefore\) Projection vector of \(\vec{a}\) on \(\vec{b} = \frac{5}{6} (\hat{i} - 2\hat{j} + \hat{k})\)

Question. Find the vector product of the vectors \(3\hat{i} - \hat{k}\) and \(-\hat{i} - \hat{j} + 5\hat{k}\).
Answer: Let \(\vec{a} = 3\hat{i} - \hat{k}\) and \(\vec{b} = -\hat{i} - \hat{j} + 5\hat{k}\)
Now, \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ -1 & -1 & 5 \end{vmatrix}\)
\(= \hat{i}(0 - 1) - \hat{j}(15 - 1) + \hat{k}(-3 - 0)\)
\(= -\hat{i} - 14\hat{j} - 3\hat{k}\)

Question. Write the value of \((\hat{k} \times \hat{j}) \cdot \hat{i} + \hat{j} \cdot \hat{k}\).
Answer: We have, \((\hat{k} \times \hat{j}) \cdot \hat{i} + \hat{j} \cdot \hat{k} = (-\hat{i}) \cdot \hat{i} + 0 = -1\)

Question. Write the angle between the vectors \(\vec{a} \times \vec{b}\) and \(\vec{b} \times \vec{a}\).
Answer: We know that \(\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\)
So, \(\vec{a} \times \vec{b}\) and \(\vec{b} \times \vec{a}\) are vectors of same magnitude but opposite in sign.
Thus, the angle between \(\vec{a} \times \vec{b}\) and \(\vec{b} \times \vec{a}\) is \(\pi\).

Question. Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be three vectors such that \(\vec{a} \neq \vec{0}\) and \(\vec{a} \times \vec{b} = 2\vec{a} \times \vec{c}\), \(|\vec{a}| = |\vec{c}| = 1\), \(|\vec{b}| = 4\) and \(|\vec{b} \times \vec{c}| = \sqrt{15}\) and \(\vec{b} - 2\vec{c} = \lambda \vec{a}\). Then, find the value of \(\lambda\).
Answer: Given, \(|\vec{a}| = |\vec{c}| = 1\) and \(|\vec{b}| = 4\)
Let angle between \(\vec{b}\) and \(\vec{c}\) be \(\alpha\).
Then, \(|\vec{b} \times \vec{c}| = \sqrt{15}\)
\(\Rightarrow |\vec{b}| |\vec{c}| \sin \alpha = \sqrt{15}\)
\(\Rightarrow 4 \times 1 \sin \alpha = \sqrt{15}\)
\(\Rightarrow \sin \alpha = \frac{\sqrt{15}}{4}\)
\(\therefore \cos \alpha = \frac{1}{4}\) \([\because \cos \theta = \sqrt{1 - \sin^2 \theta}]\)
We have, \(\vec{b} - 2\vec{c} = \lambda \vec{a}\)
\(\Rightarrow (\vec{b} - 2\vec{c})^2 = (\lambda \vec{a})^2\) [squaring both sides]
\(\Rightarrow \vec{b}^2 + 4\vec{c}^2 - 4\vec{b} \cdot \vec{c} = \lambda^2 \vec{a}^2\)
\(\Rightarrow 16 + 4 - 4 |\vec{b}| |\vec{c}| \cos \alpha = \lambda^2\)
\(\Rightarrow 20 - 16 \cos \alpha = \lambda^2\)
\(\Rightarrow 20 - 16 \times \frac{1}{4} = \lambda^2\)
\(\Rightarrow 16 = \lambda^2\)
\(\Rightarrow \lambda = \pm 4\)

Question. If \((\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2 = 225\) and \(|\vec{a}| = 5\), then find the value of \(|\vec{b}|\).
Answer: We have, \((\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2 = 225\)
\(\Rightarrow (|\vec{a}| |\vec{b}| \sin \theta)^2 + (|\vec{a}| |\vec{b}| \cos \theta)^2 = 225\)
\(\Rightarrow |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 225\)
\(\Rightarrow |\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta) = 225\)
\(\Rightarrow |\vec{a}|^2 |\vec{b}|^2 = 225\) \([\because \sin^2 A + \cos^2 A = 1]\)
\(\Rightarrow |\vec{b}|^2 = \frac{225}{25}\) \([\because |\vec{a}| = 5 \text{ (given)}]\)
\(\Rightarrow |\vec{b}|^2 = 9\)
\(\therefore |\vec{b}| = 3\)

Question. If \(\vec{a} \cdot \vec{b} = \vec{0}\) and \(\vec{a} \times \vec{b} = \vec{0}\), then prove that \(\vec{a} = \vec{0}\) or \(\vec{b} = \vec{0}\).
Answer: We have, \(\vec{a} \cdot \vec{b} = \vec{0}\) and \(\vec{a} \times \vec{b} = \vec{0}\)
\(\Rightarrow (\vec{a} = \vec{0} \text{ or } \vec{b} = \vec{0} \text{ or } \vec{a} \perp \vec{b})\) and \((\vec{a} = \vec{0} \text{ or } \vec{b} = \vec{0} \text{ or } \vec{a} \parallel \vec{b})\)
Hence, \(\vec{a} = \vec{0}\) or \(\vec{b} = \vec{0}\) \([\because \vec{a} \perp \vec{b} \text{ and } \vec{a} \parallel \vec{b} \text{ cannot occur simultaneously}]\)

Question. Find the area of the triangle whose two sides are represented by the vectors \(2\hat{i}\) and \(-3\hat{j}\).
Answer: Let \(\vec{a} = 2\hat{i}\) and \(\vec{b} = -3\hat{j}\)
\(\therefore\) Area of triangle \(= \frac{1}{2} |\vec{a} \times \vec{b}|\)
\(= \frac{1}{2} |(2\hat{i}) \times (-3\hat{j})|\)
\(= \frac{1}{2} |-6\hat{k}|\) \([\because \hat{i} \times \hat{j} = \hat{k}]\)
\(= \frac{1}{2} \times 6 = 3\) sq units

Question. Find \(\lambda\) and \(\mu\), if \((\hat{i} + 3\hat{j} + 9\hat{k}) \times (3\hat{i} - \lambda\hat{j} + \mu\hat{k}) = \vec{0}\).
Answer: Given, \((\hat{i} + 3\hat{j} + 9\hat{k}) \times (3\hat{i} - \lambda\hat{j} + \mu\hat{k}) = \vec{0}\)
\(\therefore \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 9 \\ 3 & -\lambda & \mu \end{vmatrix} = \vec{0}\)
\(\Rightarrow \hat{i}(3\mu + 9\lambda) - \hat{j}(\mu - 27) + \hat{k}(-\lambda - 9) = 0\hat{i} + 0\hat{j} + 0\hat{k}\)
On comparing the coefficients of \(\hat{i}, \hat{j}\) and \(\hat{k}\), we get
\(3\mu + 9\lambda = 0\), \(-\mu + 27 = 0\) and \(-\lambda - 9 = 0\)
\(\Rightarrow \mu = 27\) and \(\lambda = -9\)
Also, the values of \(\mu\) and \(\lambda\) satisfy the equation \(3\mu + 9\lambda = 0\).
Hence, \(\mu = 27\) and \(\lambda = -9\).

Question. Write the number of vectors of unit length perpendicular to both the vectors \(\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{j} + \hat{k}\).
Answer: We know that unit vectors perpendicular to \(\vec{a}\) and \(\vec{b}\) are \(\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\)
So, there are two unit vectors perpendicular to the given vectors.

Short Answer Type Questions

Question. Find \(|\vec{a} - \vec{b}|\), if two vectors \(\vec{a}\) and \(\vec{b}\) are such that \(|\vec{a}| = 2, |\vec{b}| = 3\) and \(\vec{a} \cdot \vec{b} = 5\).
Answer: Given, \(|\vec{a}| = 2\), \(|\vec{b}| = 3\) and \(\vec{a} \cdot \vec{b} = 5\)
Now, \(|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}\)
\(= 4 + 9 - 2 \times 5 = 13 - 10 = 3\)
\(\Rightarrow |\vec{a} - \vec{b}|^2 = 3\)
\(\therefore |\vec{a} - \vec{b}| = \sqrt{3}\)

Question. If \(\vec{a} = 3\hat{i} - \hat{j}\) and \(\vec{b} = 2\hat{i} + \hat{j} - 3\hat{k}\), then express \(\vec{b}\) in the form \(\vec{b} = \vec{b}_1 + \vec{b}_2\), where \(\vec{b}_1 \parallel \vec{a}\) and \(\vec{b}_2 \perp \vec{a}\).
Answer: Let \(\vec{b}_1 = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}\) and \(\vec{b}_2 = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}\) be two vectors such that \(\vec{b}_1 + \vec{b}_2 = \vec{b}\), where \(\vec{b}_1 \parallel \vec{a}\) and \(\vec{b}_2 \perp \vec{a}\).
Consider, \(\vec{b}_1 + \vec{b}_2 = \vec{b}\)
\(\Rightarrow (x_1 + x_2)\hat{i} + (y_1 + y_2)\hat{j} + (z_1 + z_2)\hat{k} = 2\hat{i} + \hat{j} - 3\hat{k}\)
On comparing the coefficients of \(\hat{i}, \hat{j}\) and \(\hat{k}\) both sides, we get
\(x_1 + x_2 = 2\) ...(i)
\(y_1 + y_2 = 1\) ...(ii)
and \(z_1 + z_2 = -3\) ...(iii)
Now, consider, \(\vec{b}_1 \parallel \vec{a}\)
\(\Rightarrow \frac{x_1}{3} = \frac{y_1}{-1} = \frac{z_1}{0} = \lambda\) (say)
\(\Rightarrow x_1 = 3\lambda\), \(y_1 = -\lambda\) and \(z_1 = 0\) ...(iv)
On substituting the values of \(x_1\), \(y_1\) and \(z_1\), from Eq. (iv) to Eq. (i), (ii) and (iii), respectively, we get
\(x_2 = 2 - 3\lambda\), \(y_2 = 1 + \lambda\) and \(z_2 = -3\) ...(v)
Since, \(\vec{b}_2 \perp \vec{a}\), therefore \(\vec{b}_2 \cdot \vec{a} = 0\)
\(\Rightarrow 3x_2 - y_2 = 0\)
\(\Rightarrow 3(2 - 3\lambda) - (1 + \lambda) = 0\) [from Eq. (v)]
\(\Rightarrow 6 - 9\lambda - 1 - \lambda = 0\)
\(\Rightarrow 5 - 10\lambda = 0 \Rightarrow \lambda = \frac{1}{2}\)
On substituting \(\lambda = \frac{1}{2}\) in Eqs. (iv) and (v), we get
\(x_1 = \frac{3}{2}\), \(y_1 = -\frac{1}{2}\), \(z_1 = 0\)
and \(x_2 = \frac{1}{2}\), \(y_2 = \frac{3}{2}\) and \(z_2 = -3\)
Hence, \(\vec{b}_1 = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}\) and \(\vec{b}_2 = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}\), where \(\vec{b}_1 \parallel \vec{a}\) and \(\vec{b}_2 \perp \vec{a}\).

Question. If \(\vec{a} = \hat{i} - \hat{j} + 7\hat{k}\) and \(\vec{b} = 5\hat{i} - \hat{j} + \lambda\hat{k}\), then find the value of \(\lambda\), so that \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) are perpendicular vectors.
Answer: Given, \(\vec{a} = \hat{i} - \hat{j} + 7\hat{k}\) and \(\vec{b} = 5\hat{i} - \hat{j} + \lambda\hat{k}\)
Then, \(\vec{a} + \vec{b} = (\hat{i} - \hat{j} + 7\hat{k}) + (5\hat{i} - \hat{j} + \lambda\hat{k}) = 6\hat{i} - 2\hat{j} + (7 + \lambda)\hat{k}\)
and \(\vec{a} - \vec{b} = (\hat{i} - \hat{j} + 7\hat{k}) - (5\hat{i} - \hat{j} + \lambda\hat{k}) = -4\hat{i} + (7 - \lambda)\hat{k}\)
Since, \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\) are perpendicular vectors, then
\((\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0\)
\(\Rightarrow [6\hat{i} - 2\hat{j} + (7 + \lambda)\hat{k}] \cdot [-4\hat{i} + (7 - \lambda)\hat{k}] = 0\)
\(\Rightarrow -24 + (7 + \lambda)(7 - \lambda) = 0\)
\(\Rightarrow 49 - \lambda^2 = 24\)
\(\Rightarrow \lambda^2 = 25\)
\(\therefore \lambda = \pm 5\)

Question. Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three vectors such that \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\) and \(|\vec{a}| = 1, |\vec{b}| = 2, |\vec{c}| = 3\) then find \(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}\).
Answer: We have, \(|\vec{a}| = 1, |\vec{b}| = 2\) and \(|\vec{c}| = 3\)
Now, \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\)
On squaring both sides, we get
\((\vec{a} + \vec{b} + \vec{c})^2 = (\vec{0})^2\)
\(\Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0\)
\(\Rightarrow (1)^2 + (2)^2 + (3)^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0\)
\(\Rightarrow 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -(1 + 4 + 9)\)
\(\Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{1}{2}(14)\)
\(\Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -7\)

Long Answer Type Questions

Question. If \(\vec{a} = \hat{i} + 2\hat{j} + \hat{k}\), \(\vec{b} = 2\hat{i} + \hat{j}\) and \(\vec{c} = 3\hat{i} - 4\hat{j} - 5\hat{k}\), then find a unit vector perpendicular to both of the vectors \((\vec{a} - \vec{b})\) and \((\vec{c} - \vec{b})\).
Answer: Given, vectors are \(\vec{a} = \hat{i} + 2\hat{j} + \hat{k}\), \(\vec{b} = 2\hat{i} + \hat{j}\) and \(\vec{c} = 3\hat{i} - 4\hat{j} - 5\hat{k}\)
Now, \(\vec{a} - \vec{b} = (\hat{i} + 2\hat{j} + \hat{k}) - (2\hat{i} + \hat{j}) = -\hat{i} + \hat{j} + \hat{k}\)
and \(\vec{c} - \vec{b} = (3\hat{i} - 4\hat{j} - 5\hat{k}) - (2\hat{i} + \hat{j}) = \hat{i} - 5\hat{j} - 5\hat{k}\)
Now, a vector perpendicular to \((\vec{a} - \vec{b})\) and \((\vec{c} - \vec{b})\) is given by
\((\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -5 & -5 \end{vmatrix} = \hat{i}(-5 + 5) - \hat{j}(5 - 1) + \hat{k}(5 - 1) = -4\hat{j} + 4\hat{k}\)
and unit vector along \((\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})\) is given by
\(\frac{-4\hat{j} + 4\hat{k}}{|-4\hat{j} + 4\hat{k}|} = \frac{-4\hat{j} + 4\hat{k}}{\sqrt{(-4)^2 + 4^2}} = \frac{-4\hat{j} + 4\hat{k}}{\sqrt{32}} = \frac{-4\hat{j} + 4\hat{k}}{4\sqrt{2}} = -\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\)

Question. If \(\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}\), \(\vec{b} = -\hat{i} + \hat{k}\) and \(\vec{c} = 2\hat{j} - \hat{k}\) are three vectors, then find the area of the parallelogram having diagonals \((\vec{a} + \vec{b})\) and \((\vec{b} + \vec{c})\).
Answer: Given, \(\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}\), \(\vec{b} = -\hat{i} + \hat{k}\) and \(\vec{c} = 2\hat{j} - \hat{k}\)
Let \(\vec{d}_1 = \vec{a} + \vec{b}\) and \(\vec{d}_2 = \vec{b} + \vec{c}\).
Then, \(\vec{d}_1 = (2\hat{i} - 3\hat{j} + \hat{k}) + (-\hat{i} + \hat{k}) = \hat{i} - 3\hat{j} + 2\hat{k}\)
and \(\vec{d}_2 = (-\hat{i} + \hat{k}) + (2\hat{j} - \hat{k}) = -\hat{i} + 2\hat{j}\)
Clearly, area of given parallelogram with diagonals \(\vec{d}_1\) and \(\vec{d}_2\) is given by \(\frac{1}{2} |\vec{d}_1 \times \vec{d}_2|\).
Here, \(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix} = \hat{i}(-4) - \hat{j}(0 + 2) + \hat{k}(2 - 3) = -4\hat{i} - 2\hat{j} - \hat{k}\)
So, area of parallelogram \(= \frac{1}{2} |-4\hat{i} - 2\hat{j} - \hat{k}|\)
\(= \frac{1}{2} \sqrt{(-4)^2 + (-2)^2 + (-1)^2} = \frac{1}{2} \sqrt{16 + 4 + 1} = \frac{1}{2} \sqrt{21}\) sq units

Question. Find the vector \(\vec{p}\), which is perpendicular to both \(\vec{\alpha} = 4\hat{i} + 5\hat{j} - \hat{k}\) and \(\vec{\beta} = \hat{i} - 4\hat{j} + 5\hat{k}\) and \(\vec{p} \cdot \vec{q} = 21\), where \(\vec{q} = 3\hat{i} + \hat{j} - \hat{k}\).
Answer: Given, \(\vec{\alpha} = 4\hat{i} + 5\hat{j} - \hat{k}\), \(\vec{\beta} = \hat{i} - 4\hat{j} + 5\hat{k}\) and \(\vec{q} = 3\hat{i} + \hat{j} - \hat{k}\)
Also, vector \(\vec{p}\) is perpendicular to \(\vec{\alpha}\) and \(\vec{\beta}\).
Then, \(\vec{p} = \lambda (\vec{\alpha} \times \vec{\beta})\) ...(i)
Now, \(\vec{\alpha} \times \vec{\beta} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & -1 \\ 1 & -4 & 5 \end{vmatrix} = \hat{i}(25 - 4) - \hat{j}(20 + 1) + \hat{k}(-16 - 5) = 21\hat{i} - 21\hat{j} - 21\hat{k}\)
So, \(\vec{p} = 21\lambda\hat{i} - 21\lambda\hat{j} - 21\lambda\hat{k}\) [from Eq. (i)] ...(ii)
Also, given that \(\vec{p} \cdot \vec{q} = 21\)
\(\Rightarrow (21\lambda\hat{i} - 21\lambda\hat{j} - 21\lambda\hat{k}) \cdot (3\hat{i} + \hat{j} - \hat{k}) = 21\)
\(\Rightarrow 63\lambda - 21\lambda + 21\lambda = 21\)
\(\Rightarrow 63\lambda = 21 \Rightarrow \lambda = \frac{1}{3}\)
On putting \(\lambda = \frac{1}{3}\) in Eq. (ii), we get
\(\vec{p} = 21 \times \frac{1}{3}\hat{i} - 21 \times \frac{1}{3}\hat{j} - 21 \times \frac{1}{3}\hat{k}\)
\(\therefore \vec{p} = 7\hat{i} - 7\hat{j} - 7\hat{k}\), which is the required vector.

Question. Find the unit vector perpendicular to the plane ABC, where the position vectors of A, B and C are \(2\hat{i} - \hat{j} + \hat{k}\), \(\hat{i} + \hat{j} + 2\hat{k}\) and \(2\hat{i} + 3\hat{k}\), respectively.
Answer: Let O be the origin of reference.
Then, \(\vec{OA} = 2\hat{i} - \hat{j} + \hat{k}\), \(\vec{OB} = \hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{OC} = 2\hat{i} + 3\hat{k}\)
Now, \(\vec{AB} = \vec{OB} - \vec{OA} = (\hat{i} + \hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} + 2\hat{j} + \hat{k}\)
and \(\vec{AC} = \vec{OC} - \vec{OA} = (2\hat{i} + 3\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = \hat{j} + 2\hat{k}\)
Now, \(\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix} = \hat{i}(4 - 1) - \hat{j}(-2 - 0) + \hat{k}(-1 - 0) = 3\hat{i} + 2\hat{j} - \hat{k}\)
and \(|\vec{AB} \times \vec{AC}| = \sqrt{(3)^2 + (2)^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}\)
\(\dots\) Unit vector perpendicular to the plane ABC \(= \frac{\vec{AB} \times \vec{AC}}{|\vec{AB} \times \vec{AC}|} = \frac{3\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{14}} = \frac{3}{\sqrt{14}}\hat{i} + \frac{2}{\sqrt{14}}\hat{j} - \frac{1}{\sqrt{14}}\hat{k}\)

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: Given, \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{b} = \hat{j} - \hat{k}\)
Let \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\) ...(i)
Now, \(\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{i}(z - y) - \hat{j}(z - x) + \hat{k}(y - x)\)
Now, \(\vec{a} \times \vec{c} = \vec{b}\) [given]
\(\Rightarrow \hat{i}(z - y) + \hat{j}(x - z) + \hat{k}(y - x) = 0\hat{i} + 1\hat{j} + (-1)\hat{k}\) \([\because \vec{b} = \hat{j} - \hat{k}]\)
On comparing the coefficients from both sides, we get
\(z - y = 0\), \(x - z = 1\), \(y - x = -1\)
\(\Rightarrow y = z\) and \(x - y = 1\) ...(ii)
Also given, \(\vec{a} \cdot \vec{c} = 3\)
\(\Rightarrow (\hat{i} + \hat{j} + \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 3\)
\(\Rightarrow x + y + z = 3\)
\(\Rightarrow x + 2y = 3\) \([\because y = z]\) ...(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
\(3y = 2 \Rightarrow y = \frac{2}{3} = z\) \([\because y = z]\)
From Eq. (ii),
\(x = 1 + y = 1 + \frac{2}{3} = \frac{5}{3}\)
Hence, \(\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\) [from Eq. (i)]