NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Miscellaneous Exercise

Miscellaneous Exercise

1. Write down a unit vector in the XY-plane, making an angle of ${30^\circ }$ with the positive direction of the x-axis. 

Ans: Unit vector is $\vec r = \cos \theta \hat i + \sin \theta \hat j$, where $\theta $ is angle with positive ${\text{X}}$ axis. $\vec r = \cos {30^\circ }\hat i + \sin {30^\circ }\hat j = \dfrac{{\sqrt 3 }}{2}\hat i + \dfrac{1}{2}\hat j$

2. Find the scalar components and magnitude of the vector joining the points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$

Ans: $\overrightarrow {{\text{PQ}}}  = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k$

$|\overrightarrow {{\text{PQ}}} |\; = \;\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $

3. A girl walks 4 km towards the west, then she walks 3 km in a direction ${30^\circ }$ east of north and stops. Determine the girl's displacement from her initial point of departure. 

Ans:

$\overrightarrow {{\text{OA}}}  =  - 4\hat i$

$\overrightarrow {{\text{AB}}}  = \hat i|\overrightarrow {{\text{AB}}} |\cos {60^\circ } + \hat j|\overrightarrow {{\text{AB}}} |\sin {60^\circ }$

$ = \hat i3 \times \dfrac{1}{2} + \hat j3 \times \dfrac{{\sqrt 3 }}{2}$

$ = \dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$

$\overrightarrow {{\text{OB}}}  = \overrightarrow {{\text{OA}}}  + \overrightarrow {{\text{AB}}} $

$ = ( - 4\hat i) + \left( {\dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}j} \right)$

$ = \left( { - 4 + \dfrac{3}{2}} \right)\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$

$ = \left( {\dfrac{{ - 8 + 3}}{9}} \right)\hat i + \dfrac{{3\sqrt 3 }}{9}\hat j$

$ = \dfrac{{ - 5}}{2}\vec i + \dfrac{{3\sqrt 3 }}{2}\hat j$

4. If $\overrightarrow a  = \vec b + \vec c$, then is it true that $|\overrightarrow a | = |\vec b| + |\vec c|$ ? Justify your answer. 

Ans:  $\operatorname{In} \Delta {\text{ABC}},\overrightarrow {\;CB}  = \vec a,\;\overrightarrow {CA}  = b,\;\overrightarrow {AB}  = \vec c$

$\vec a = \vec b + \vec c$,  by triangle law of addition for vectors.

$|\vec a|\; < \;|\vec b| + |\vec c|$,  by triangle inequality law of lengths.

 Hence, it's not true that $|\vec a| = |\vec b| + |\vec c|$

5. Find the value of $x$ for which $x(\hat i + \hat j + \hat k)$ unit vector.

Ans: $|x(\hat i + \hat j + \hat k)| = 1$

$ \Rightarrow \sqrt {{x^2} + {x^2} + {x^2}}  = 1$

$ \Rightarrow \sqrt {3{x^2}}  = 1$

$ \Rightarrow \sqrt 3 x = 1$

$ \Rightarrow x =  \pm \dfrac{1}{{\sqrt 3 }}$

6.  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 $\vec b = \hat i - 2\hat j + \hat k$

Ans: $\vec c = \vec a + \vec b = (2 + 1)\hat i + (3 - 2)\hat j + ( - 1 + 1)\hat k = 3\hat i + \hat j$

$|\vec c| = \sqrt {{3^2} + {1^2}}  = \sqrt {9 + 1}  = \sqrt {10} $

$\therefore \hat c = \dfrac{{\vec c}}{{|\vec c|}} = \dfrac{{(3\hat i + \hat j)}}{{\sqrt {10} }}$

So, a vector of magnitude 5 and parallel to the resultant of $\vec a$ and $\vec b$ is $ \pm 5(\hat c) =  \pm 5\left( {\dfrac{1}{{\sqrt {10} }}(3\hat i + \hat j)} \right) =  \pm \dfrac{{3\sqrt {10} }}{2}\hat i \pm \dfrac{{\sqrt {10} }}{2}\hat j$

7.  If $\vec a = \hat i + \hat j + \hat k\;,\;\vec b = 2\hat i - \hat j + 3\hat k$ and $\vec c = \hat i - 2\hat j + \hat k$, find a unit vector parallel to the vector $2\vec a - \vec b + 3\vec c$.

Ans: $2\vec a - \vec b + 3\vec c = 2(\hat i + \hat j + \hat k) - (2\hat i - \hat j + 3\hat k) + 3(\hat i - 2\hat j + \hat k)$

$ = 2\hat i + 2\hat j + 2\hat k - 2\hat i + j - 3\vec k + 3\hat i - 6\hat j + 3\hat k$

$ = 3\hat i - 3\hat j + 2\hat k$

$|2\vec a - \vec b + 3c| = \sqrt {{3^2} + {{( - 3)}^2} + {2^2}}  = \sqrt {9 + 9 + 4}  = \sqrt {22} $

Thus ,required unit vector is $\dfrac{{2\vec a - \vec b + 3\vec c}}{{|2\vec a - \vec b + 3\vec c|}} = \dfrac{{3\hat i - 3\hat j + 2\hat k}}{{\sqrt {22} }} = \dfrac{3}{{\sqrt {22} }}\hat i - \dfrac{3}{{\sqrt {22} }}\hat j + \dfrac{2}{{\sqrt {22} }}\hat k$

8. Show that the points ${\text{A}}(1, - 2, - 8),{\text{B}}(5,0, - 2)$ and ${\text{C}}(11,3,7)$ are collinear, and find the ratio in which B divides Ac. 

Ans: $\overrightarrow {{\text{AB}}}  = (5 - 1)\hat i + (0 + 2)\hat j + ( - 2 + 8)\hat k = 4\hat i + 2\hat j + 6\hat k$

$\overrightarrow {{\text{BC}}}  = (11 - 5)\hat i + (3 - 0)\hat j + (7 + 2)\hat k = 6\hat i + 3\hat j + 9\hat k$

$\overrightarrow {AC}  = (11 - 1)\hat i + (3 + 2)\hat j + (7 + 8)\hat k = 10\hat i + 5\hat j + 15\hat k$

$|\overrightarrow {{\text{AB}}} | = \sqrt {{4^2} + {2^2} + {6^2}}  = \sqrt {16 + 4 + 36}  = \sqrt {56}  = 2\sqrt {14} $

$|\overrightarrow {{\text{BC}}} | = \sqrt {{6^2} + {3^2} + {9^2}}  = \sqrt {36 + 9 + 81}  = \sqrt {126}  = 3\sqrt {14} $

$|\overrightarrow {AC} | = \sqrt {{{10}^2} + {5^2} + {{15}^2}}  = \sqrt {100 + 25 + 225}  = \sqrt {350}  = 5\sqrt {14} $

$\therefore \;\;|\overrightarrow {{\text{AC}}} | = |\overrightarrow {{\text{AB}}} | + |\overrightarrow {{\text{BC}}} |$

So, the points are collinear.

Let B divide AC in ratio $\lambda :1$. $\overrightarrow {{\text{OB}}}  = \dfrac{{\lambda \overrightarrow {{\text{OC}}}  + \overrightarrow {{\text{OA}}} }}{{(\lambda  + 1)}}$

$ \Rightarrow 5\hat i - 2\hat k = \dfrac{{\lambda (11\hat i + 3\hat j + 7\hat k) + (\hat i - 2\hat j - 8\hat k)}}{{\lambda  + 1}}$

$ \Rightarrow (\lambda  + 1)(5\hat i - 2\hat k) = 11\lambda \hat i + 3\lambda \hat j + 7\lambda \hat k + \hat i - 2\hat j - 8\hat k$

$ \Rightarrow 5(\lambda  + 1)\hat i - 2(\lambda  + 1)\hat k = (11\lambda  + 1)\hat i + (3\lambda  - 2)\hat j + (7\lambda  - 8)\hat k$

$ \Rightarrow \lambda  = \dfrac{2}{3}$

So, the required ratio is 2:3

9. Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec a + \vec b)$ and $(\vec a - 3\vec b)$ externally in the ratio 1: 2. Also, show that $P$ is the midpoint of the line segment RQ. 

Ans: $\overrightarrow {{\text{OP}}}  = 2\vec a + \vec b,\overrightarrow {{\text{OQ}}}  = \overrightarrow {\text{a}}  - 3\overrightarrow {\text{b}} $

$\overrightarrow {OR}  = \dfrac{{2(2\vec a + \vec b) - (\vec a - 3\vec b)}}{{2 - 1}} = \dfrac{{4\vec a + 2\vec b - \vec a - 3\vec b}}{1} = 3\vec a + 5\vec b$

So, the position vector of ${\text{R}}$ is $3\overrightarrow {\text{a}}  + 5\overrightarrow {\text{b}} $

Position vector of midpoint of ${\text{RQ}} = \dfrac{{\overrightarrow {{\text{OQ}}}  + \overrightarrow {{\text{OR}}} }}{2}$

$ = \dfrac{{(a\sqrt 6 ) + (3\vec a + 5\bar b)}}{2}$

$ = 2\vec a + \vec b$

$ = \overrightarrow {OP} $

Thus, ${\text{P}}$ is midpoint of line segment ${\text{RQ}}$

10. 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 its diagonal. Also, find its area. 

Ans:  Diagonal of a parallelogram is $\vec a + \vec b$ 

$\vec a + \vec b = (2 + 1)\hat i + ( - 4 - 2)\hat j + (5 - 3)\hat k = 3\hat i - 6\hat j + 2\hat k$

So, the unit vector parallel to diagonal is $\dfrac{{\vec a + \vec b}}{{|\vec a + \vec b|}} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {{3^2} + {{( - 6)}^2} + {2^2}} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {9 + 36 + 4} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{7} = \dfrac{3}{7}\hat i - \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k$

$\vec{a}\times \vec{b} = \begin{vmatrix} \hat i& \hat j &\hat k \\ 2& -4 & 3\\ 1& -2 & -3 \end{vmatrix}$

$ = \hat i(12 + 10) - \hat j( - 6 - 5) + \bar k( - 4 + 4)$

$ = 22\hat i + 11\hat j$

$ = 11(2\hat i + \hat j)$

$\therefore |\vec a \times \vec b| = 11\sqrt {{2^2} + {1^2}}  = 11\sqrt 5 $

So, the area of the parallelogram is $11\sqrt 5 $ sq units

11. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ  are $\dfrac{1}{{\sqrt 3 }}$ $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.

Ans: Let a vector be equally inclined to ${\text{OX}},{\text{OY}}$, and ${\text{OZ}}$ at an angle $\alpha $.

So, the Direction Cosines of the vector are $\cos \alpha ,\cos \alpha $and $\cos \alpha $.

${\cos ^2}\alpha  + {\cos ^2}\alpha  + {\cos ^2}\alpha  = 1$

$ \Rightarrow 3{\cos ^2}\alpha  = 1$

$ \Rightarrow \cos \alpha  = \dfrac{1}{{\sqrt 3 }}$

So, the DCs of the vector are $\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}$.

12. Let $\vec a = \hat i + 4\hat j + 2\hat k$ and $\vec b = 3\hat i - 2\hat j + 7\hat k$ 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 $\vec c \cdot \vec d = 15$ 

Ans: $\vec d = {d_1}\hat i + {d_2}\hat \jmath  + {d_3}\hat k$

$\vec d\;.\;\vec a = 0 \Rightarrow {d_1} + 4{d_2} + 2{d_3} = 0$

$\vec d\;.\;\vec b = 0 \Rightarrow 3{d_1} - 2{d_2} + 7{d_3} = 0$

$\vec c \cdot \vec d = 15 \Rightarrow 2{d_1} - {d_2} + 4{d_3} = 15$

Solving these equations, we get ${d_1} = \dfrac{{160}}{3},{d_2} =  - \dfrac{5}{3},{d_3} =  - \dfrac{{70}}{3}$

$\therefore \vec d = \dfrac{{160}}{3}\hat i + \dfrac{5}{3}\widehat j + \dfrac{{70}}{3}\hat k = \dfrac{1}{3}(160\hat i + 5\hat j + 70\hat k)$

13. The scalar product of the vector $\hat i + \hat j + \hat k$ with a unit vector along the sum of vectors $2\hat i + 4\hat j - 5\hat k$ änd $\lambda \hat i + 2\hat j + 3\hat k$ is equal to one. Find the value of $\lambda $. 

Ans: $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k) = (2 + \lambda )\hat i + 6\hat j - 2\hat k$

So, unit vector along $(2\hat i + 4\hat j - 5\hat k) + (\lambda \hat i + 2\hat j + 3\hat k)$ is $\left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }}} \right)$

$(\hat i + \hat j + \hat k) \cdot \left( {\dfrac{{(2 + \lambda )\hat i + 6\hat j - 2\hat k}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }}} \right) = 1$

$ \Rightarrow \dfrac{{(2 + \lambda ) + 6 - 2}}{{\sqrt {{\lambda ^2} + 4\lambda  + 44} }} = 1$

$ \Rightarrow \sqrt {{\lambda ^2} + 4\lambda  + 44}  = \lambda  + 6$

$ \Rightarrow {\lambda ^2} + 4\lambda  + 44 = {(\lambda  + 6)^2}$

$ \Rightarrow {\lambda ^2} + 4\lambda  + 44 = {\lambda ^2} + 12\lambda  + 36$

$ \Rightarrow 8\lambda  = 8$

$ \Rightarrow \lambda  = 1$

14. 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 $\vec c$. 

Ans: $\vec a\;.\;\vec b = \vec b\;.\;\vec c = \vec c\;.\;\vec a = 0$

$|\vec a| = |\vec b| = |\vec c|$

Let $\vec a + \vec b + \vec c$ be inclined to $\vec a,\vec b,\vec c$ at angles ${\theta _1},{\theta _2},{\theta _3}$ respectively. $\cos {\theta _1} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{\vec a \cdot \vec a + \vec b \cdot \vec a + \vec c \cdot \vec a}}{{|\vec a + \vec b + \vec c||\vec a\mid }} = \dfrac{{|\vec a{|^2}}}{{|\vec a + \vec b + \vec c||\vec a|}} = \dfrac{{|\vec a|}}{{|\vec a + \vec b + \vec c|}}$

$\cos {\theta _2} = \dfrac{{(\vec a + \vec b + \vec c)\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{\vec a\vec b + \vec b\vec b + \vec c\vec b}}{{|\vec a + \vec b + \vec c||\vec b\mid }} = \dfrac{{|\vec b{|^2}}}{{|\vec a + \vec b + \vec c||\vec b|}} = \dfrac{{|\vec b|}}{{|\vec a + \vec b + \vec c|}}$

$\operatorname{os} {\theta _3} = \dfrac{{(\vec a + \vec b + \vec c) \cdot \vec c}}{{|\vec a + \vec b + \vec c||\bar c|}} = \dfrac{{\vec a\vec c + \vec b\vec c + \vec c\vec c}}{{|\vec a + \vec b + \vec c||\vec c\mid }} = \dfrac{{|\vec c{|^2}}}{{|\vec a + \vec b + \vec c||\vec c|}} = \dfrac{{|\vec c|}}{{|\vec a + \vec b + \vec c|}}$

Since, $|\vec a| = |\vec b| = |\vec c|\; \Rightarrow \;\cos {\theta _1} = \cos {\theta _2} = \cos {\theta _3}$, So, ${\theta _1} = {\theta _2} = {\theta _3}$

15. Prove that, $(\vec a + \vec b) \cdot (\vec a + \vec b) = \;|\vec a{|^2} + |\vec b{|^2}$ if and only if $\vec a,\vec b$ are perpendicular, given $\vec a \ne \vec 0,\vec b \ne 0$

Ans: $(\vec a + \vec b) \cdot (\vec a + \vec b) = |\vec a{|^2} + |\vec b{|^2}$

$ \Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b \cdot \vec a + \vec b\;.\;\vec b = |\vec a{|^2} + |\vec b{|^2}$

$ \Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\vec b{|^2} = |a{|^2} + |\vec b{|^2}$

$ \Rightarrow 2\vec a\;.\;\vec b = 0$

$ \Rightarrow \vec a\;.\;\vec b = 0$

So $\vec a$ and $\vec b$ are perpendicular.

16. If $\theta $ is the angle between two vectors $\vec a$ and $\vec b$, then $\vec a\vec b \geqslant 0$ only when

a. $0 < \theta  < \dfrac{\pi }{2}$

b. $0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

c. $0 < \theta  < \pi $

d. $0 \leqslant \theta  \leqslant \pi $

Ans: $\therefore \vec a\vec b \geqslant 0$

$ \Rightarrow |\vec a||\vec b|\cos \theta  \geqslant 0$

$ \Rightarrow \cos \theta  \geqslant 0\quad \because [|\vec a| \geqslant 0$ and $|\vec b| \geqslant 0]$

$ \Rightarrow 0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

$\vec a \cdot \vec b \geqslant 0$ if $0 \leqslant \theta  \leqslant \dfrac{\pi }{2}$

So the right answer is B

17. Let $\vec a$ and $\vec b$ be two unit vectors and $\theta $ is the angle between them. Then $\vec a + \vec b$ is a unit vector if

a. $\theta  = \dfrac{\pi }{4}$

b. $\theta  = \dfrac{\pi }{3}$

c. $\theta  = \dfrac{\pi }{2}$

d. $\theta  = \dfrac{{2\pi }}{3}$

Ans: $|\vec a| = |\vec b| = 1$

$|\vec a + \vec b| = 1$

$ \Rightarrow (\vec a + \overrightarrow b )(\vec a + \overrightarrow b ) = 1$

$ \Rightarrow \vec a \cdot \vec a + \vec a\;.\;\vec b + \vec b\;.\;\vec a + \vec b\;.\;\vec b = 1$

$ \Rightarrow |\vec a{|^2} + 2\vec a\;.\;\vec b + |\overrightarrow b {|^2} = 1$

$ \Rightarrow {1^2} + 2|\vec a||\vec b|\cos \theta  + {1^2} = 1$

$ \Rightarrow 1 + 2.1.1\cos \theta  + 1 = 1$

$ \Rightarrow \cos \theta  =  - \dfrac{1}{2}$

$ \Rightarrow \theta  = \dfrac{{2\pi }}{3}$

So, $\vec a + \vec b$ is unit vector if $\theta  = \dfrac{{2\pi }}{3}$ 

The correct answer is D

18. The value of $\hat i.(\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$ is

$\begin{array}{*{20}{l}} {{\text{ a. }}0}&{{\text{ b. }} - 1}&{{\text{ c. }}1}&{{\text{ d. }}3} \end{array}$

Ans:

$\hat i \cdot (\hat j \times \hat k) + \hat j \cdot (\hat i \times \hat k) + \hat k \cdot (\hat i \times \hat j)$

$ = \hat i\;.\;\hat i + \hat j \cdot ( - \hat j) + \hat k\;.\;\hat k$

$ = 1 - 1 + 1$

$ = 1$

The correct answer is C.

19. If $\theta $ is the angle between any two vectors $\vec a$ and $\vec b$, then $|\vec a\vec b| = |\vec a \times \vec b|$ when $\theta $ is equal to

a. 0 

b. $\dfrac{\pi }{4}$

c. $\dfrac{\pi }{2}$

d. n

Ans: $|\vec a\vec b| = |\vec a \times \vec b|$

$ \Rightarrow |\vec a|\vec b|\cos \theta  = |\vec a||\vec b\mid \sin \theta $

$ \Rightarrow \cos \theta  = \sin \theta $

$ \Rightarrow \tan \theta  = 1$

$ \Rightarrow \theta  = \dfrac{\pi }{4}$

The correct answer is B

Conclusion

Miscellaneous Exercise Class 12 Chapter 10 is important for understanding various concepts thoroughly. Class 12 Maths Chapter 10 Miscellaneous Exercise Solutions covers diverse problems that require the application of multiple formulas and techniques. Rather than simply learning answers by heart, it's important to concentrate on understanding the fundamental ideas behind each question.

Class 12 Maths Chapter 10: Exercises Breakdown

CBSE Class 12 Maths Chapter 10 Other Study Materials

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.