CBSE Class 12 Mathematics Vector Algebra VBQs Set B

Very Short Answer Questions 

Question. Find a vector in the direction of vector \( \vec{a} = \hat{i} - 2\hat{j} \) that has magnitude 7 units. 
Answer: The unit vector in the direction of the given vector \( \vec{a} \) is
\( \hat{a} = \frac{1}{|\vec{a}|} \vec{a} = \frac{1}{\sqrt{5}} (\hat{i} - 2\hat{j}) = \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j} \)
Therefore, the vector having magnitude equal to 7 and in the direction of \( \vec{a} \) is
\( 7\hat{a} = 7 \left( \frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j} \right) = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j} \)

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: Number of vectors of unit length perpendicular to both vectors = 2

Question. Write the value of \( p \) for which \( \vec{a} = 3\hat{i} + 2\hat{j} + 9\hat{k} \) and \( \vec{b} = \hat{i} + p\hat{j} + 3\hat{k} \) are parallel vector. 
Answer: Since \( \vec{a} \parallel \vec{b} \), therefore \( \vec{a} = \lambda\vec{b} \Rightarrow 3\hat{i} + 2\hat{j} + 9\hat{k} = \lambda(\hat{i} + p\hat{j} + 3\hat{k}) \)
\( \Rightarrow \lambda = 3, 2 = \lambda p, 9 = 3\lambda \text{ or } \lambda = 3, p = \frac{2}{3} \) [By comparing the coefficients]

Question. Write a vector of magnitude 15 units in the direction of vector \( \hat{i} - 2\hat{j} + 2\hat{k} \).
Answer: Let \( \vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \)
Unit vector in the direction of \( \vec{a} \) is \( \hat{a} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{\sqrt{(1)^2 + (-2)^2 + (2)^2}} = \frac{1}{3}(\hat{i} - 2\hat{j} + 2\hat{k}) \)
Vector of magnitude 15 units in the direction of \( \vec{a} = 15 \hat{a} = 15 \frac{(\hat{i} - 2\hat{j} + 2\hat{k})}{3} = 5\hat{i} - 10\hat{j} + 10\hat{k} \)

Question. What is the cosine of the angle, which the vector \( \sqrt{2}\hat{i} + \hat{j} + \hat{k} \) makes with y-axis?
Answer: We will consider \( \vec{a} = \sqrt{2}\hat{i} + \hat{j} + \hat{k} \)
Unit vector in the direction of \( \vec{a} \) is \( \hat{a} = \frac{\sqrt{2}\hat{i} + \hat{j} + \hat{k}}{\sqrt{(\sqrt{2})^2 + (1)^2 + (1)^2}} = \frac{\sqrt{2}\hat{i} + \hat{j} + \hat{k}}{\sqrt{4}} = \frac{\sqrt{2}\hat{i} + \hat{j} + \hat{k}}{2} \)
\( = \frac{\sqrt{2}}{2}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} \)
The cosine of the angle which the vector \( \sqrt{2}\hat{i} + \hat{j} + \hat{k} \) makes with y-axis is \( \left( \frac{1}{2} \right) \).

Question. If \( |\vec{a}| = 4, |\vec{b}| = 3 \) and \( \vec{a} \cdot \vec{b} = 6\sqrt{3} \), then the value of \( |\vec{a} \times \vec{b}| \). 
Answer: We have, \( \vec{a} \cdot \vec{b} = 6\sqrt{3} \Rightarrow |\vec{a}| |\vec{b}| \cos \theta = 6\sqrt{3} \)
\( \Rightarrow 4 \times 3 \cos \theta = 6\sqrt{3} \Rightarrow \cos \theta = \frac{6\sqrt{3}}{4 \times 3} = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{6} \)
Now, \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 4 \times 3 \sin \frac{\pi}{6} = 4 \times 3 \times \frac{1}{2} = 6 \)

Question. Write the value of the area of the parallelogram determined by the vectors \( 2\hat{i} \) and \( 3\hat{j} \). 
Answer: Required area of parallelogram \( = |2\hat{i} \times 3\hat{j}| = 6 |\hat{i} \times \hat{j}| = 6 |\hat{k}| = 6 \) sq units.
[Note: Area of parallelogram whose sides are represented by \( \vec{a} \) and \( \vec{b} \) is \( |\vec{a} \times \vec{b}| \)]

Question. Write the value of \( (\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i} \). 
Answer: \( (\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i} = \hat{k} \cdot \hat{k} + \hat{i} \cdot \hat{i} = 1 + 1 = 2 \)
[Note: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \). Also \( |\hat{i}| = |\hat{j}| = |\hat{k}| = 1 \)]

Question. For what value of ‘a’ the vectors \( 2\hat{i} - 3\hat{j} + 4\hat{k} \) and \( a\hat{i} + 6\hat{j} - 8\hat{k} \) are collinear? 
Answer: \( \because 2\hat{i} - 3\hat{j} + 4\hat{k} \) and \( a\hat{i} + 6\hat{j} - 8\hat{k} \) are collinear
\( \therefore \frac{2}{a} = \frac{-3}{6} = \frac{4}{-8} \Rightarrow a = \frac{2 \times 6}{-3} \text{ or } a = \frac{2 \times (-8)}{4} \Rightarrow a = -4 \)
[Note: If \( \vec{a} \) and \( \vec{b} \) are collinear vectors then the respective components of \( \vec{a} \) and \( \vec{b} \) are proportional.]

Question. Write the direction cosines of the vector \( -2\hat{i} + \hat{j} - 5\hat{k} \). 
Answer: Direction cosines of vector \( -2\hat{i} + \hat{j} - 5\hat{k} \) are
\( \frac{-2}{\sqrt{(-2)^2 + 1^2 + (-5)^2}}, \frac{1}{\sqrt{(-2)^2 + 1^2 + (-5)^2}}, \frac{-5}{\sqrt{(-2)^2 + 1^2 + (-5)^2}} = \frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}} \)
Note: If \( l, m, n \) are direction cosine of \( a\hat{i} + b\hat{j} + c\hat{k} \) then
\( l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} \)

Question. If vectors \( \vec{a} \) and \( \vec{b} \) are such that \( |\vec{a}| = \frac{1}{2}, |\vec{b}| = \frac{4}{\sqrt{3}} \) and \( |\vec{a} \times \vec{b}| = \frac{1}{\sqrt{3}} \), then find \( |\vec{a} \cdot \vec{b}| \).
Answer: We have, \( |\vec{a} \times \vec{b}| = \frac{1}{\sqrt{3}} \Rightarrow |\vec{a}| |\vec{b}| \sin \theta = \frac{1}{\sqrt{3}} \)
\( \Rightarrow \frac{1}{2} \times \frac{4}{\sqrt{3}} \sin \theta = \frac{1}{\sqrt{3}} \Rightarrow \sin \theta = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{1}{2} \Rightarrow \theta = 30^\circ \)
Now, \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = \frac{1}{2} \times \frac{4}{\sqrt{3}} \cos 30^\circ = \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = 1 \)

Question. Let \( \vec{a} \) and \( \vec{b} \) be two vectors such that \( |\vec{a}| = 3 \) and \( |\vec{b}| = \frac{\sqrt{2}}{3} \) and \( \vec{a} \times \vec{b} \) is a unit vector. What is the angle between \( \vec{a} \) and \( \vec{b} \)?
Answer: We have, \( |\vec{a}| = 3, |\vec{b}| = \frac{\sqrt{2}}{3} \)
Now, \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \Rightarrow 1 = 3 \times \frac{\sqrt{2}}{3} \sin \theta \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = 45^\circ \)

Question. Give an example of vectors \( \vec{a} \) and \( \vec{b} \) such that \( |\vec{a}| = |\vec{b}| \) but \( \vec{a} \neq \vec{b} \). 
Answer: Let \( \vec{a} = x\hat{i} + y\hat{j}; \vec{b} = y\hat{i} + x\hat{j} \)
\( |\vec{a}| = \sqrt{x^2 + y^2}, |\vec{b}| = \sqrt{y^2 + x^2} \). Hence, \( \vec{a} \neq \vec{b} \) but \( |\vec{a}| = |\vec{b}| \)

Question. Find the unit vector in the direction of sum of vectors \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{j} + \hat{k} \). 
Answer: Let \( \vec{c} \) denote the sum of \( \vec{a} \) and \( \vec{b} \)
We have, \( \vec{c} = \vec{a} + \vec{b} = 2\hat{i} - \hat{j} + \hat{k} + 2\hat{j} + \hat{k} = 2\hat{i} + \hat{j} + 2\hat{k} \)
\( \therefore \) Unit vector in the direction of \( \vec{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{9}} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} \)

Short Answer Questions-

Question. Show that for any two non-zero vectors \( \vec{a} \) and \( \vec{b}, |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \) iff \( \vec{a} \) and \( \vec{b} \) are perpendicular vectors.
Answer: We have, \( |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \)
\( \Leftrightarrow |\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \) (Squaring both sides)
\( \Leftrightarrow |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}| |\vec{b}| \cos \theta = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}| |\vec{b}| \cos \theta \)
(Where \( \theta \) is the angle between vectors \( \vec{a} \) and \( \vec{b} \))
\( \Leftrightarrow 4|\vec{a}| |\vec{b}| \cos \theta = 0 \Rightarrow \cos \theta = 0 = \cos \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{2} \)
\( \Leftrightarrow \vec{a} \) and \( \vec{b} \) are perpendicular vectors.

Question. The x-coordinate of a point on the line joining the point P(2, 2, 1) and Q(5, 1, –2) is 4. Find its z-coordinate. 
Answer: Let required point be R(4, \( y_1, z_1 \)) which divides PQ in ratio \( k : 1 \).
By section formula, \( 4 = \frac{5k + 2}{k + 1} \Rightarrow 4k + 4 = 5k + 2 \Rightarrow k = 2 \)
\( \therefore z_1 = \frac{2 \times (-2) + 1 \times 1}{2 + 1} = \frac{-4 + 1}{3} = \frac{-3}{3} = -1 \)

Question. Find ‘\( \lambda \)’ when the projection of \( \vec{a} = \lambda\hat{i} + \hat{j} + 4\hat{k} \) on \( \vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k} \) is 4 units. 
Answer: We know that projection of \( \vec{a} \) on \( \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \Rightarrow 4 = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \)...(i)
Now, \( \vec{a} \cdot \vec{b} = 2\lambda + 6 + 12 = 2\lambda + 18 \) also \( |\vec{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = 7 \)
Putting in (i), we get \( 4 = \frac{2\lambda + 18}{7} \Rightarrow 2\lambda = 28 - 18 \Rightarrow \lambda = \frac{10}{2} = 5 \)

Question. What are the direction cosines of a line, which makes equal angles with the co-ordinate axes? 
Answer: Let \( \alpha \) be the angle made by line with coordinate axes.
\( \Rightarrow \) Direction cosines of line are \( \cos \alpha, \cos \alpha, \cos \alpha \).
\( \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \Rightarrow 3 \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = \frac{1}{3} \Rightarrow \cos \alpha = \pm \frac{1}{\sqrt{3}} \)
Hence, the direction cosines of the line equally inclined to the coordinate axes are \( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \)
[Note: If \( l, m, n \) are direction cosines of line then \( l^2 + m^2 + n^2 = 1 \)]

Question. For what value of \( p \), is \( (\hat{i} + \hat{j} + \hat{k})p \) a unit vector? 
Answer: Let, \( \vec{a} = p(\hat{i} + \hat{j} + \hat{k}) \)
Magnitude of \( \vec{a} = |\vec{a}| = \sqrt{(p)^2 + (p)^2 + (p)^2} = \pm \sqrt{3}p \)
As \( \vec{a} \) is a unit vector, \( |\vec{a}| = 1 \Rightarrow \pm \sqrt{3}p = 1 \Rightarrow p = \pm \frac{1}{\sqrt{3}} \)

Question. X and Y are two points with position vectors \( 3\vec{a} + \vec{b} \) and \( \vec{a} - 3\vec{b} \) respectively. Write the position vector of a point Z which divides the line segment XY in the ratio 2 : 1 externally. [CBSE 2019 (65/4/1)]
Answer: We have \( \vec{OX} = 3\vec{a} + \vec{b}, \vec{OY} = \vec{a} - 3\vec{b}, \vec{OZ} = ? \)
\( \vec{OZ} = \frac{2(\vec{a} - 3\vec{b}) - 1(3\vec{a} + \vec{b})}{2 - 1} = \frac{-\vec{a} - 7\vec{b}}{1} = -\vec{a} - 7\vec{b} \)
\( \therefore \vec{OZ} = -\vec{a} - 7\vec{b} \)

Question. Write the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2). 
Answer: Let \( \vec{a}, \vec{b} \) be position vector of points P(2, 3, 4) and Q(4, 1, –2) respectively.
\( \therefore \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{b} = 4\hat{i} + \hat{j} - 2\hat{k} \)
\( \therefore \) Position vector of mid point of P and Q \( = \frac{\vec{a} + \vec{b}}{2} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k} \)

Question. If \( |\vec{a}| = a \), then find the value of the following : \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 \). 
Answer: Let \( \vec{a} \) makes angle \( \alpha, \beta, \gamma \) with x, y and z axis.
\( \therefore |\vec{a} \times \hat{i}| = |\vec{a}| \cdot 1 \cdot \sin \alpha = a \sin \alpha \). Similarly \( |\vec{a} \times \hat{j}| = a \sin \beta \) and \( |\vec{a} \times \hat{k}| = a \sin \gamma \)
\( \therefore |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = a^2 \sin^2 \alpha + a^2 \sin^2 \beta + a^2 \sin^2 \gamma = a^2 [\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma] \)
\( = a^2 [1 - \cos^2 \alpha + 1 - \cos^2 \beta + 1 - \cos^2 \gamma] = a^2 [3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma)] \)
\( = a^2 (3 - 1) = 2a^2 \quad [\because l^2 + m^2 + n^2 = 1 \Rightarrow \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1] \)

Question. The vectors \( \vec{a} = 3\hat{i} + x\hat{j} \) and \( \vec{b} = 2\hat{i} + \hat{j} + y\hat{k} \) are mutually perpendicular. If \( |\vec{a}| = |\vec{b}| \), then find the value of y. 
Answer: \( \because \vec{a} \) and \( \vec{b} \) are mutually perpendicular.
\( \therefore \vec{a} \cdot \vec{b} = 0 \Rightarrow (3\hat{i} + x\hat{j}) \cdot (2\hat{i} + \hat{j} + y\hat{k}) = 0 \Rightarrow 6 + x + 0 \cdot y = 0 \Rightarrow 6 + x = 0 \Rightarrow x = -6 \)
Again, \( |\vec{a}| = |\vec{b}| \Rightarrow \sqrt{3^2 + x^2} = \sqrt{2^2 + 1^2 + y^2} \Rightarrow \sqrt{9 + 36} = \sqrt{5 + y^2} \quad [\because x = -6] \)
\( \Rightarrow 45 = 5 + y^2 \Rightarrow y^2 = 40 \Rightarrow y = \pm \sqrt{40} = \pm 2\sqrt{10} \)

Question. Find the value of \( \vec{a} \cdot \vec{b} \) if \( |\vec{a}| = 10, |\vec{b}| = 2 \) and \( |\vec{a} \times \vec{b}| = 16 \). 
Answer: \( \because |\vec{a} \times \vec{b}| = 16 \Rightarrow |\vec{a}| |\vec{b}| \sin \theta = 16 \Rightarrow 10 \times 2 \sin \theta = 16 \Rightarrow \sin \theta = \frac{16}{20} = \frac{4}{5} \)
\( \Rightarrow \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{16}{25}} = \pm \frac{3}{5} \)
\( \therefore \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 10 \times 2 \times \left( \pm \frac{3}{5} \right) = \pm 12 \)

Question. Find the vector of magnitude 6, which is perpendicular to both the vectors \( 2\hat{i} - \hat{j} + 2\hat{k} \) and \( 4\hat{i} - \hat{j} + 3\hat{k} \). 
Answer: Let \( \vec{a} = 2\hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{b} = 4\hat{i} - \hat{j} + 3\hat{k} \)
So, any vector perpendicular to both the vectors \( \vec{a} \) and \( \vec{b} \) is given by
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{vmatrix} = \hat{i}(-3 + 2) - \hat{j}(6 - 8) + \hat{k}(-2 + 4) = -\hat{i} + 2\hat{j} + 2\hat{k} = \vec{r} \text{ [say]} \)
A vector of magnitude 6 in the direction of \( \vec{r} \) is \( 6 \cdot \frac{\vec{r}}{|\vec{r}|} = 6 \cdot \frac{-\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{6}{3}(-\hat{i} + 2\hat{j} + 2\hat{k}) = -2\hat{i} + 4\hat{j} + 4\hat{k} \)

Question. Let \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \) be two vectors. Show that the vectors \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) are perpendicular to each other. 
Answer: We have \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \).
Then, \( \vec{a} + \vec{b} = 4\hat{i} + \hat{j} - \hat{k} \) and \( \vec{a} - \vec{b} = -2\hat{i} + 3\hat{j} - 5\hat{k} \)
\( \because (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = -8 + 3 + 5 = 0 \Rightarrow (\vec{a} + \vec{b}) \perp (\vec{a} - \vec{b}) \)

Short Answer Questions-

Question. For any two vectors \( \vec{a} \) and \( \vec{b} \), prove that \( (\vec{a} \times \vec{b})^2 = \vec{a}^2 \vec{b}^2 - (\vec{a} \cdot \vec{b})^2 \). 
Answer: \( \because (\vec{a} \times \vec{b}) = |\vec{a}| |\vec{b}| \sin \theta \hat{n} \text{ and } \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \)
RHS \( = \vec{a}^2 \vec{b}^2 - (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = |\vec{a}|^2 |\vec{b}|^2 (1 - \cos^2 \theta) = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta = (\vec{a} \times \vec{b})^2 = \text{LHS} \)

Question. The scalar product of the vector \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) with a unit vector along the sum of the vectors \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k} \) is equal to 1. Find the value of \( \lambda \) and hence find the unit vector along \( \vec{b} + \vec{c} \). 
Answer: We have, \( \vec{b} + \vec{c} = (2\hat{i} + 4\hat{j} - 5\hat{k}) + (\lambda\hat{i} + 2\hat{j} + 3\hat{k}) \Rightarrow \vec{b} + \vec{c} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \)
\( \therefore \) Unit vector of \( \vec{b} + \vec{c} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 36 + 4}} \)
Now, \( \vec{a} \cdot \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} = 1 \Rightarrow (\hat{i} + \hat{j} + \hat{k}) \cdot \frac{((2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k})}{\sqrt{(2 + \lambda)^2 + 40}} = 1 \)
\( \Rightarrow 2 + \lambda + 6 - 2 = \sqrt{(2 + \lambda)^2 + 40} \Rightarrow \lambda + 6 = \sqrt{(2 + \lambda)^2 + 40} \)
Squaring both sides, we have \( \lambda^2 + 12\lambda + 36 = (2 + \lambda)^2 + 40 = 4 + \lambda^2 + 4\lambda + 40 \)
\( \Rightarrow 8\lambda = 44 - 36 = 8 \Rightarrow \lambda = 1 \)
By putting the value of \( \lambda = 1 \), unit vector of \( \vec{b} + \vec{c} = \frac{3}{7}\hat{i} + \frac{6}{7}\hat{j} - \frac{2}{7}\hat{k} \)

Question. Let \( \vec{a}, \vec{b} \) and \( \vec{c} \) be three vectors such that \( |\vec{a}| = 1, |\vec{b}| = 2 \) and \( |\vec{c}| = 3 \). If the projection of \( \vec{b} \) along \( \vec{a} \) is equal to the projection of \( \vec{c} \) along \( \vec{a} \); and \( \vec{b}, \vec{c} \) are perpendicular to each other, then find \( |3\vec{a} - 2\vec{b} + 2\vec{c}| \). 
Answer: Given projection of \( \vec{b} \) along \( \vec{a} \) is equal to the projection of \( \vec{c} \) along \( \vec{a} \).
\( \Rightarrow \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} = \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} \Rightarrow \vec{b} \cdot \vec{a} = \vec{c} \cdot \vec{a} \)...(i)
Also given \( \vec{b} \perp \vec{c} \Rightarrow \vec{b} \cdot \vec{c} = 0 \)...(ii)
Now, \( |3\vec{a} - 2\vec{b} + 2\vec{c}|^2 = 9|\vec{a}|^2 + 4|\vec{b}|^2 + 4|\vec{c}|^2 - 12\vec{a} \cdot \vec{b} - 8\vec{b} \cdot \vec{c} + 12\vec{a} \cdot \vec{c} \)
\( = 9(1)^2 + 4(2)^2 + 4(3)^2 - 12\vec{a} \cdot \vec{b} - 8(0) + 12\vec{a} \cdot \vec{b} = 9 + 16 + 36 = 61 \)
\( \Rightarrow |3\vec{a} - 2\vec{b} + 2\vec{c}| = \sqrt{61} \)

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

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

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 \( \Rightarrow [\vec{a}, \vec{b}, \vec{c}] = 0 \). Now, \( [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = 2[\vec{a}, \vec{b}, \vec{c}] = 2 \times 0 = 0 \). 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 \( \Rightarrow [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = 0 \Rightarrow 2[\vec{a}, \vec{b}, \vec{c}] = 0 \Rightarrow [\vec{a}, \vec{b}, \vec{c}] = 0 \). 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: (a) Since \( \vec{a}, \vec{b} \) and \( \vec{c} \) vectors are coplanar \( \Rightarrow \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & c_3 \end{vmatrix} = 0 \Rightarrow 1(0 - 0) - 1(c_3 - 0) + 1(2 - 0) = 0 \Rightarrow -c_3 + 2 = 0 \Rightarrow c_3 = 2 \).
(b) To make \( \vec{a}, \vec{b} \) and \( \vec{c} \) coplanar \( \Rightarrow \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ c_1 & -1 & 1 \end{vmatrix} = 0 \Rightarrow 1(0 - 0) - 1(1 - 0) + 1(-1 - 0) = 0 \Rightarrow -1 - 1 = 0 \Rightarrow -2 = 0 \), which is never possible. Hence, if \( c_2 = -1 \) and \( c_3 = 1 \), there is no value of \( c_1 \) which can make \( \vec{a}, \vec{b} \) 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 \( \vec{a}, \vec{b} \) or \( \vec{c} \). 
Answer: Let \( |\vec{a}| = |\vec{b}| = |\vec{c}| = x \). Since \( \vec{a}, \vec{b}, \vec{c} \) are mutually perpendicular, \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0 \).
Now, \( |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(0) = 3x^2 \Rightarrow |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}x \).
Let \( \theta_1, \theta_2, \theta_3 \) be the angles. \( \cos \theta_1 = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{a}| |\vec{a} + \vec{b} + \vec{c}|} = \frac{x^2}{x \cdot \sqrt{3}x} = \frac{1}{\sqrt{3}} \).
\( \theta_1 = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \), similarly \( \theta_2 = \theta_3 = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \).

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 \( \vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \). \( \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ c_1 & c_2 & c_3 \end{vmatrix} = (c_3 - c_2)\hat{i} + (c_1 - c_3)\hat{j} + (c_2 - c_1)\hat{k} \).
Given \( \vec{a} \times \vec{c} = \vec{b} \Rightarrow (c_3 - c_2)\hat{i} + (c_1 - c_3)\hat{j} + (c_2 - c_1)\hat{k} = 0\hat{i} + \hat{j} - \hat{k} \).
\( c_3 - c_2 = 0, c_1 - c_3 = 1, c_2 - c_1 = -1 \)...(i).
Also, \( \vec{a} \cdot \vec{c} = 3 \Rightarrow c_1 + c_2 + c_3 = 3 \)...(ii).
Solving, \( c_1 = \frac{5}{3}, c_2 = \frac{2}{3}, c_3 = \frac{2}{3} \). Hence, \( \vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \).

Question. If \( \vec{a} + \vec{b} + \vec{c} = 0 \) 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: \( \vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow (\vec{a} + \vec{b})^2 = (-\vec{c})^2 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{c}|^2 \)
\( \Rightarrow 9 + 25 + 2|\vec{a}| |\vec{b}| \cos \theta = 49 \Rightarrow 34 + 30 \cos \theta = 49 \Rightarrow 30 \cos \theta = 15 \Rightarrow \cos \theta = \frac{1}{2} = \cos 60^\circ \Rightarrow \theta = 60^\circ \).

Question. If \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \( \vec{a} + \vec{b} + \vec{c} = 0 \), then prove that \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \), and hence show that \( [\vec{a} \vec{b} \vec{c}] = 0 \). 
Answer: \( \vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow \vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times 0 \Rightarrow 0 + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = 0 \Rightarrow \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \). Similarly, \( \vec{b} \times \vec{c} = \vec{a} \times \vec{b} \). Hence, \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \). Also \( [\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (\vec{a} \times \vec{b}) = 0 \).

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: Vector \( \vec{p} \) is parallel to \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = 32\hat{i} - \hat{j} - 14\hat{k} \).
Let \( \vec{p} = \mu(32\hat{i} - \hat{j} - 14\hat{k}) \). Given \( \vec{p} \cdot \vec{c} = 18 \Rightarrow \mu(32\hat{i} - \hat{j} - 14\hat{k}) \cdot (2\hat{i} - \hat{j} + 4\hat{k}) = 18 \)
\( \Rightarrow \mu(64 + 1 - 56) = 18 \Rightarrow 9\mu = 18 \Rightarrow \mu = 2 \).
Hence, \( \vec{p} = 2(32\hat{i} - \hat{j} - 14\hat{k}) = 64\hat{i} - 2\hat{j} - 28\hat{k} \).

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: \( \left| \vec{a} \times \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} \right| = \sqrt{2} \). \( \vec{b} + \vec{c} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \).
\( |\vec{b} + \vec{c}| = \sqrt{(2 + \lambda)^2 + 40} \). \( \vec{a} \times (\vec{b} + \vec{c}) = -8\hat{i} + (4 + \lambda)\hat{j} + (4 - \lambda)\hat{k} \).
Putting in equation: \( \frac{\sqrt{(-8)^2 + (4 + \lambda)^2 + (4 - \lambda)^2}}{\sqrt{\lambda^2 + 4\lambda + 44}} = \sqrt{2} \).
Squaring: \( \frac{64 + 16 + \lambda^2 + 8\lambda + 16 + \lambda^2 - 8\lambda}{\lambda^2 + 4\lambda + 44} = 2 \Rightarrow \frac{96 + 2\lambda^2}{\lambda^2 + 4\lambda + 44} = 2 \Rightarrow 8\lambda = 8 \Rightarrow \lambda = 1 \).