CBSE Class 12 Mathematics Vector Algebra Previous Year Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Find the value of 'p' for which the vectors \( 3\hat{i} + 2\hat{j} + 9\hat{k} \) and \( \hat{i} - 2p\hat{j} + 3\hat{k} \) are parallel. [CBSE 2014, 1M]
Answer:
Two vectors are parallel if \( \frac{3}{1} = \frac{2}{-2p} = \frac{9}{3} \)
Then \( p = -\frac{1}{3} \) [1]

Question 2. Find a vector \( \vec{a} \) of magnitude \( 5\sqrt{2} \), making an angle of \( \frac{\pi}{4} \) with X-axis, \( \frac{\pi}{2} \) with Y-axis and an acute angle θ with Z-axis. [CBSE 2014, 1M]
Answer:
Here, \( l = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( m = \cos\frac{\pi}{2} = 0 \)
Therefore, \( l^2 + m^2 + n^2 = 1 \) gives
\( \frac{1}{2} + 0 + n^2 = 1 \Rightarrow n = \frac{1}{\sqrt{2}} \) (θ is acute, so n > 0)
Hence, the required vector \( \vec{r} = 5\sqrt{2}(l\hat{i} + m\hat{j} + n\hat{k}) \) is given by
\( \vec{r} = 5\sqrt{2}\left(\frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}\right) \Rightarrow \vec{r} = 5\hat{i} + 5\hat{k} \) ...(1) [1]

Question 3. In a triangle OAC, if B is the mid-point of side AC and \( \overrightarrow{OA} = \vec{a} \), \( \overrightarrow{OB} = \vec{b} \), then what is \( \overrightarrow{OC} \)? [CBSE 2015, 1M]
Answer:
\( \overrightarrow{OB} = \frac{\overrightarrow{OA} + \overrightarrow{OC}{2}} \)
⇒ \( \vec{b} = \frac{\vec{a} + \overrightarrow{OC}{2}} \)
or \( \overrightarrow{OC} = 2\vec{b} - \vec{a} \) [1]

Question 4. Find the vector of magnitude \( \sqrt{171} \) which is perpendicular to both of the vectors \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \). [CBSE 2015, 1M]
Answer:
Let the required vector be \( \vec{c} = \lambda(\vec{a} \times \vec{b}) \)
⇒ \( \vec{c} = \lambda\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 3 & -1 & 2 \end{vmatrix} \)
or \( \vec{c} = \lambda(\hat{i} - 11\hat{j} - 7\hat{k}) \) ...(1)
⇒ \( |\vec{c}| = |\lambda|\sqrt{1 + 121 + 49} \)
or \( \sqrt{171} = |\lambda|\sqrt{171} \) (given)
⇒ λ = ±1; putting in equation (1), we have
\( \vec{c} = \pm(\hat{i} - 11\hat{j} - 7\hat{k}) \) [1]

Question 5. If \( \vec{a} = 4\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k} \), then find a unit vector parallel to the vector \( \vec{a} + \vec{b} \). [CBSE 2016, 1M]
Answer:
\( \vec{a} + \vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k} \)
∴ Required unit vector \( = \pm\frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \pm\frac{(6\hat{i} - 3\hat{j} + 2\hat{k})}{\sqrt{36 + 9 + 4}} = \pm\frac{(6\hat{i} - 3\hat{j} + 2\hat{k})}{7} \) [1]

Question 6. Find λ and μ if \( (\hat{i} + 3\hat{j} + 9\hat{k}) \times (3\hat{i} - \lambda\hat{j} + \mu\hat{k}) = \vec{0} \). [CBSE 2016, 1M]
Answer:
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 9 \\ 3 & -\lambda & \mu \end{vmatrix} = \vec{0} \) (given)
⇒ \( \hat{i}(3\mu + 9\lambda) - \hat{j}(\mu - 27) + \hat{k}(-\lambda - 9) = \vec{0} \)
∴ μ − 27 = 0 ...(i), −λ − 9 = 0 ...(ii), 3μ + 9λ = 0 ...(iii)
μ = 27, λ = −9, which are satisfied by eq. (iii) [½ + ½]

Question 7. Find the magnitude of each of the two vectors \( \vec{a} \) and \( \vec{b} \), having the same magnitude, such that the angle between them is 60° and their scalar product is \( \frac{9}{2} \). [CBSE 2018, 1M]
Answer:
Given, \( |\vec{a}| = |\vec{b}| \)
Angle between \( \vec{a} \) and \( \vec{b} \) is 60°
and \( \vec{a} \cdot \vec{b} = \frac{9}{2} \)
⇒ \( |\vec{a}||\vec{b}|\cos 60^\circ = \frac{9}{2} \)
or \( |\vec{a}|^2 \times \frac{1}{2} = \frac{9}{2} \) \( (\because |\vec{a}| = |\vec{b}|) \)
⇒ \( |\vec{a}|^2 = 9 \)
or \( |\vec{a}| = 3 \)
Hence, the magnitudes of \( \vec{a} \) and \( \vec{b} \) are equal to 3. [1]

Question 8. If \( |\vec{a}| = 4 \) and \( -3 \leq \lambda \leq 2 \), then \( |\lambda\vec{a}| \) lies in [CBSE 2020, 1M]
(A) [0, 12]    (B) [2, 3]    (C) [8, 12]    (D) [−12, 8]
Answer:
Ans. (A)
\( |\vec{a}| = 4 \) and \( -3 \leq \lambda \leq 2 \)
Now \( |\lambda\vec{a}| = |\lambda||\vec{a}| \)
⇒ \( |\lambda\vec{a}| = 4|\lambda| \) \( (\because |\vec{a}| = 4) \)
∵ \( -3 \leq \lambda \leq 2 \)
For λ = −3 ⇒ \( |\lambda\vec{a}| = 12 \) and for λ = 2 ⇒ \( |\lambda\vec{a}| = 8 \)
∵ \( |\lambda\vec{a}| \geq 0 \) (it is 0 when λ = 0)
∴ \( |\lambda\vec{a}| \in [0, 12] \) [1]

Question 9. The area of a triangle formed by vertices O, A and B, where \( \overrightarrow{OA} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \overrightarrow{OB} = -3\hat{i} - 2\hat{j} + \hat{k} \) is [CBSE 2020, 1M]
(A) \( 3\sqrt{5} \) sq. units    (B) \( 5\sqrt{5} \) sq. units    (C) \( 6\sqrt{5} \) sq. units    (D) 4 sq. units
Answer:
Ans. (A)
\( \overrightarrow{OA} = \hat{i} + 2\hat{j} + 3\hat{k} \), \( \overrightarrow{OB} = -3\hat{i} - 2\hat{j} + \hat{k} \)
Area of ΔOAB = \( \frac{1}{2}|\overrightarrow{OA} \times \overrightarrow{OB}| \)
Now \( \overrightarrow{OA} \times \overrightarrow{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{vmatrix} = 8\hat{i} - 10\hat{j} + 4\hat{k} \)
⇒ \( |\overrightarrow{OA} \times \overrightarrow{OB}| = \sqrt{64 + 100 + 16} = \sqrt{180} \)
So area of ΔOAB = \( \frac{1}{2}\sqrt{180} = 3\sqrt{5} \) sq. units [1]

Short Answer Questions (2 Marks)

Question 10. If θ is the angle between two vectors \( \hat{i} - 2\hat{j} + 3\hat{k} \) and \( 3\hat{i} - 2\hat{j} + \hat{k} \), find sin θ. [CBSE 2018, 2M]
Answer:
Let \( \vec{a} = (\hat{i} - 2\hat{j} + 3\hat{k}) \) and \( \vec{b} = (3\hat{i} - 2\hat{j} + \hat{k}) \)
∴ \( \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \)
\( = \frac{(\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} + \hat{k})}{\sqrt{(1)^2 + (-2)^2 + (3)^2} \times \sqrt{(3)^2 + (-2)^2 + (1)^2}} \)
\( = \frac{3 + 4 + 3}{\sqrt{14} \cdot \sqrt{14}} \)
\( = \frac{10}{14} = \frac{5}{7} \) [1]
Hence, \( \cos\theta = \frac{5}{7} \)
⇒ \( \sin\theta = \sqrt{1 - \cos^2\theta} \)
\( = \sqrt{1 - \frac{25}{49}} \)
Hence, \( \sin\theta = \sqrt{\frac{24}{49}} = \frac{2}{7}\sqrt{6} \) [1]

Question 11. Show that the points \( A(-2\hat{i} + 3\hat{j} + 5\hat{k}) \), \( B(\hat{i} + 2\hat{j} + 3\hat{k}) \) and \( C(7\hat{i} - \hat{k}) \) are collinear. [CBSE 2019, 2M]
OR
Find \( |\vec{a} \times \vec{b}| \), if \( \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k} \).
Answer:
\( \overrightarrow{AB} = 3\hat{i} - \hat{j} - 2\hat{k} \),
\( \overrightarrow{AC} = 9\hat{i} - 3\hat{j} - 6\hat{k} = 3(3\hat{i} - \hat{j} - 2\hat{k}) \) [1]
⇒ \( \overrightarrow{AC} = 3\overrightarrow{AB} \)
∴ \( \overrightarrow{AC} \parallel \overrightarrow{AB} \) but A is common in \( \overrightarrow{AC} \) and \( \overrightarrow{AB} \)
∴ A, B and C are collinear [1]
OR
\( \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k} \)
⇒ \( (\vec{a} \times \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix} = \hat{i}(-2 - 15) - \hat{j}(-4 - 9) + \hat{k}(10 - 3) \)
⇒ \( (\vec{a} \times \vec{b}) = -17\hat{i} + 13\hat{j} + 7\hat{k} \) [1½]
or \( |\vec{a} \times \vec{b}| = \sqrt{(-17)^2 + (13)^2 + (7)^2} \)
\( = \sqrt{289 + 169 + 49} = \sqrt{507} \)
\( = 13\sqrt{3} \) [½]

Question 12. Find a vector \( \vec{r} \) equally inclined to the three axes and whose magnitude is \( 3\sqrt{3} \) units. [CBSE 2020, 2M]
OR
Find the angle between unit vectors \( \vec{a} \) and \( \vec{b} \) so that \( \sqrt{3}\vec{a} - \vec{b} \) is also a unit vector.
Answer:
Let α be the angle made by the vector \( \vec{r} \) with the coordinate axes.
Then \( \vec{r} = 3\sqrt{3}(l\hat{i} + m\hat{j} + n\hat{k}) \) \( (\because |\vec{r}| = 3\sqrt{3}) \)
where l = cos α, m = cos α, n = cos α
∵ \( l^2 + m^2 + n^2 = 1 \)
⇒ \( 3\cos^2\alpha = 1 \Rightarrow \cos\alpha = \pm\frac{1}{\sqrt{3}} \) [1]
∴ \( \vec{r} = 3\sqrt{3}\left(\pm\frac{\hat{i}}{\sqrt{3}} \pm \frac{\hat{j}}{\sqrt{3}} \pm \frac{\hat{k}}{\sqrt{3}}\right) \)
or \( \vec{r} = \pm 3(\hat{i} + \hat{j} + \hat{k}) \) [1]
OR
Given \( \vec{a} \) and \( \vec{b} \) are unit vectors, i.e. \( |\vec{a}| = 1 = |\vec{b}| \)
and \( |\sqrt{3}\vec{a} - \vec{b}| = 1 \Rightarrow |\sqrt{3}\vec{a} - \vec{b}|^2 = 1 \)
⇒ \( (\sqrt{3}\vec{a} - \vec{b}) \cdot (\sqrt{3}\vec{a} - \vec{b}) = 1 \)
or \( 3|\vec{a}|^2 - \sqrt{3}\vec{a} \cdot \vec{b} - \sqrt{3}\vec{b} \cdot \vec{a} + |\vec{b}|^2 = 1 \) \( (\because \vec{a} \cdot \vec{a} = |\vec{a}|^2) \)
⇒ \( 3 - 2\sqrt{3}\vec{a} \cdot \vec{b} + 1 = 1 \) \( (\because \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}) \) [1]
or \( \vec{a} \cdot \vec{b} = \frac{\sqrt{3}}{2} \Rightarrow |\vec{a}||\vec{b}|\cos\theta = \frac{\sqrt{3}}{2} \) \( (\because \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta) \)
⇒ \( \cos\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{6} \) [1]

Long Answer Questions I (4 Marks)

Question 13. 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 λ. [CBSE 2009, 4M]
Answer:
\( \vec{s} = (2\hat{i} + 4\hat{j} - 5\hat{k}) + (\lambda\hat{i} + 2\hat{j} + 3\hat{k}) \)
\( \vec{s} = \hat{i}(2 + \lambda) + 6\hat{j} - 2\hat{k} \)
Unit vector of \( \vec{s} = \hat{s} = \frac{\vec{s}}{|\vec{s}|} = \frac{(\lambda + 2)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(\lambda + 2)^2 + 6^2 + 2^2}} \) [1½]
Given: \( (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{s} = 1 \)
∴ \( \frac{(\lambda + 2) + 6 - 2}{\sqrt{(\lambda + 2)^2 + 40}} = 1 \) [1]
or \( (\lambda + 6) = \sqrt{(\lambda + 2)^2 + 40} \)
\( \lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44 \) (on squaring both sides)
⇒ 8λ = 8 ⇒ λ = 1 [1½]

Question 14. 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 \( \vec{p} \cdot \vec{c} = 18 \). [CBSE 2010, 2012, 4M]
Answer:
Given \( \vec{a} = \hat{i} + 4\hat{j} + 2\hat{k} \), \( \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k} \), \( \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \) and \( \vec{p} \cdot \vec{c} = 18 \)
The vector which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) must be parallel to \( \vec{a} \times \vec{b} \)
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} \)
\( = \hat{i}(28 + 4) - \hat{j}(7 - 6) + \hat{k}(-2 - 12) = 32\hat{i} - \hat{j} - 14\hat{k} \) [1]
Since \( \vec{p} \) is ⊥ to both \( \vec{a} \) and \( \vec{b} \),
i.e. \( \vec{p} \parallel (\vec{a} \times \vec{b}) \) [1]
Let \( \vec{p} = \lambda(\vec{a} \times \vec{b}) = \lambda(32\hat{i} - \hat{j} - 14\hat{k}) \)
\( \vec{p} \cdot \vec{c} = 18 \Rightarrow \lambda(32\hat{i} - \hat{j} - 14\hat{k}) \cdot (2\hat{i} - \hat{j} + 4\hat{k}) = 18 \) [½]
Then 2(32λ) + (−1)(−λ) + 4(−14λ) = 18
⇒ 64λ + λ − 56λ = 18
or 9λ = 18 ⇒ λ = 2 [½]
∴ \( \vec{p} = 2(32\hat{i} - \hat{j} - 14\hat{k}) \) [1]

Question 15. 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 \( \vec{c} \cdot \vec{d} = 27 \). [CBSE 2015, 4M]
Answer:
Given \( \vec{a} = \hat{i} + 4\hat{j} + 2\hat{k} \), \( \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k} \), \( \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \) and \( \vec{c} \cdot \vec{d} = 27 \)
The vector which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) must be parallel to \( \vec{a} \times \vec{b} \)
⇒ \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} \)
\( = \hat{i}(28 + 4) - \hat{j}(7 - 6) + \hat{k}(-2 - 12) = 32\hat{i} - \hat{j} - 14\hat{k} \) [1]
Since \( \vec{d} \) is ⊥ to both \( \vec{a} \) and \( \vec{b} \),
i.e. \( \vec{d} \parallel (\vec{a} \times \vec{b}) \) [1]
Let \( \vec{d} = \lambda(\vec{a} \times \vec{b}) = \lambda(32\hat{i} - \hat{j} - 14\hat{k}) \)
∵ \( \vec{c} \cdot \vec{d} = 27 \Rightarrow \lambda(32\hat{i} - \hat{j} - 14\hat{k}) \cdot (2\hat{i} - \hat{j} + 4\hat{k}) = 27 \) [½]
Then 2(32λ) + (−1)(−λ) + 4(−14λ) = 27
⇒ 64λ + λ − 56λ = 27
or 9λ = 27 ⇒ λ = 3 [½]
∴ \( \vec{d} = 3(32\hat{i} - \hat{j} - 14\hat{k}) \) [1]

Question 16. 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. [CBSE 2017, 4M]
Answer:
\( \overrightarrow{AB} = -\hat{i} - 2\hat{j} - 6\hat{k} \)
⇒ \( |\overrightarrow{AB}|^2 = (1^2 + 2^2 + 6^2) = 41 \)
\( \overrightarrow{BC} = 2\hat{i} - \hat{j} + \hat{k} \)
⇒ \( |\overrightarrow{BC}|^2 = (2^2 + 1^2 + 1^2) = 6 \) [1]
\( \overrightarrow{CA} = -\hat{i} + 3\hat{j} + 5\hat{k} \)
⇒ \( |\overrightarrow{CA}|^2 = (1^2 + 3^2 + 5^2) = 35 \)
∵ \( \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec{0} \)
and \( |\overrightarrow{AB}|^2 = |\overrightarrow{BC}|^2 + |\overrightarrow{CA}|^2 \)
Hence, A, B, C are the vertices of a right-angled triangle. [1]
Now, area of \( \Delta = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| \), where \( \overrightarrow{AC} = \hat{i} - 3\hat{j} - 5\hat{k} \)
\( \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -2 & -6 \\ 1 & -3 & -5 \end{vmatrix} \)
\( = (-8\hat{i} - 11\hat{j} + 5\hat{k}) \) [1]
Now \( \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}|-8\hat{i} - 11\hat{j} + 5\hat{k}| \)
\( = \frac{1}{2}\sqrt{64 + 121 + 25} = \frac{1}{2}\sqrt{210} \) sq. units [1]

Question 17. Let \( \vec{a} = 4\hat{i} + 5\hat{j} - \hat{k} \), \( \vec{b} = \hat{i} - 4\hat{j} + 5\hat{k} \) and \( \vec{c} = 3\hat{i} + \hat{j} - \hat{k} \). Find a vector \( \vec{d} \) which is perpendicular to both \( \vec{c} \) and \( \vec{b} \) and \( \vec{d} \cdot \vec{a} = 21 \). [CBSE 2018, 4M]
Answer:
Given \( \vec{a} = 4\hat{i} + 5\hat{j} - \hat{k} \), \( \vec{b} = \hat{i} - 4\hat{j} + 5\hat{k} \), \( \vec{c} = 3\hat{i} + \hat{j} - \hat{k} \) and \( \vec{d} \cdot \vec{a} = 21 \).
The vector which is perpendicular to both \( \vec{c} \) and \( \vec{b} \) must be parallel to \( \vec{c} \times \vec{b} \)
⇒ \( \vec{c} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -1 \\ 1 & -4 & 5 \end{vmatrix} = \hat{i}(5 - 4) - \hat{j}(15 + 1) + \hat{k}(-12 - 1) \)
or \( \vec{c} \times \vec{b} = \hat{i} - 16\hat{j} - 13\hat{k} \) [1]
Now \( \vec{d} \) is ⊥ to both \( \vec{c} \) and \( \vec{b} \)
i.e. \( \vec{d} \parallel \vec{c} \times \vec{b} \) or \( \vec{d} = \lambda(\vec{c} \times \vec{b}) \) [1]
\( \vec{d} = \lambda(\hat{i} - 16\hat{j} - 13\hat{k}) = (\lambda\hat{i} - 16\lambda\hat{j} - 13\lambda\hat{k}) \) ...(i)
Now \( \vec{d} \cdot \vec{a} = 21 \)
\( (\lambda\hat{i} - 16\lambda\hat{j} - 13\lambda\hat{k}) \cdot (4\hat{i} + 5\hat{j} - \hat{k}) = 21 \)
4λ − 80λ + 13λ = 21
⇒ −63λ = 21 [1]
or \( \lambda = \frac{21}{-63} = -\frac{1}{3} \)
Putting this value of λ in equation (i),
⇒ \( \vec{d} = -\frac{1}{3}(\hat{i} - 16\hat{j} - 13\hat{k}) \) [1]
or \( \vec{d} = -\frac{1}{3}\hat{i} + \frac{16}{3}\hat{j} + \frac{13}{3}\hat{k} \)

Question 18. 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 λ and hence find the unit vector along \( \vec{b} + \vec{c} \). [CBSE 2019, 4M]
Answer:
\( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k} \)
⇒ \( \vec{b} + \vec{c} = \hat{i}(2 + \lambda) + 6\hat{j} - 2\hat{k} = \vec{d} \) (let)
Unit vector along \( \vec{d} = \frac{\hat{i}(2 + \lambda) + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2}} \)
∵ \( \vec{a} \cdot \hat{d} = 1 \)
⇒ \( (\hat{i} + \hat{j} + \hat{k}) \cdot \left(\frac{\hat{i}(2 + \lambda) + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2}}\right) = 1 \) [1]
⇒ \( (2 + \lambda) + 6 - 2 = \sqrt{(2 + \lambda)^2 + 36 + 4} \) [1½]
or \( 6 + \lambda = \sqrt{4 + 4\lambda + \lambda^2 + 40} \)
⇒ \( 36 + \lambda^2 + 12\lambda = 44 + 4\lambda + \lambda^2 \)
or 8λ = 8
⇒ λ = 1 [½]
∴ Unit vector along \( (\vec{b} + \vec{c}) \)
\( = \frac{\hat{i}(2 + \lambda) + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2}} \)
\( = \frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{9 + 36 + 4}} = \frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{7} \) [1]