# NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.2) Exercise 10.2

Class 12 Chapter 10 Vector Algebra includes the concepts of Vectors, Scalars, etc. Furthermore, the chapter also includes Direction Cosines and Direction Ratios of a Vector, Types of Vectors (Zero, Unit, Equal, Parallel and Collinear Vectors), Components of a Vector, Negative of a Vector, the Addition of Vectors, Multiplication of a Vector by a Scalar, Position Vector of a Point Dividing a Line Segment in a Given Ratio, etc. Students can use the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 as a quick reference to help them understand difficult topics.

Chapter-10 Vector Algebra is not only a chapter of the Class 12 Mathematics curriculum, but it is a concept practised in other specified fields of the subject. The concept is widely used in the fields of Physics and Engineering, particularly in gravitational fields, electromagnetic fields, and fluid flow. Vector Algebra is used to determine the component of force in a particular direction. Furthermore, Vector Algebra also has multiple practical applications, such as air and water navigation. Therefore, students need to learn Vector Algebra properly in Class 12 as the students who have opted for Mathematics in Class 12 are looking forward to learning it for further studies. The Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 are the most important resources for the preparation for the Mathematics Class 12 board examination.

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## NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.2) Exercise 10.2

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### Concepts covered in Exercise 10.2:

Ex 10.2 Class 12 Maths includes the Addition of Vectors, Properties of Vector Addition, Multiplication of Vector by a Scalar, Components of a Vector, Vector Joining Two Parts and Section Formula. The Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 provided by Extramarks are very useful at the time of quick revisions. Students will have difficulty clearing their doubts if the solutions are not explained in proper steps. Also in Mathematics, students can score marks based on step-wise marking. Extramarks provides students with the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 which are properly detailed step-by-step solutions.

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### Importance of Vector Algebra

Vector Algebra deals with  quantities that involve both magnitude and direction. It involves various mathematical operations that can be applied to the vectors. Furthermore, there is a list of formulas and operations that are included in the chapter. Students can review the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 for a better understanding of the chapter.

### Applications of Vector Algebra

Vector Algebra is a concept that is widely used in the subjects of Physics and Engineering. The fields of fluid Flow, Electromagnetic Fields and Gravitational Fields are the ones in which the concept of Vector Algebra is dominantly used. Furthermore, it is also used during navigation by air and navigation by boat. Vectors also have many practical applications, specifically in situations that involve force and velocity. For a better understanding of the chapter, students can refer to the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 provided by Extramarks.

### NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.2

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Q.1

$\begin{array}{l}\text{Compute the magnitude of the following vectors:}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}};\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-7\stackrel{^}{\mathrm{j}}-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}};\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{j}}-\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{k}};\end{array}$

Ans

$\begin{array}{l}\text{The given vectors are:}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}};\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}-7\stackrel{^}{\mathrm{j}}-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}};\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{j}}-\frac{1}{\sqrt{3}}\stackrel{^}{\mathrm{k}}\\ \text{Magnitude of}\stackrel{\to }{\mathrm{a}}=|\stackrel{\to }{\mathrm{a}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}+{\left(1\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{3}\\ \text{Magnitude of}\stackrel{\to }{\mathrm{b}}=|\stackrel{\to }{\mathrm{b}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(2\right)}^{2}+{\left(-7\right)}^{2}+{\left(-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{62}\\ \text{Magnitude of}\stackrel{\to }{\mathrm{c}}=|\stackrel{\to }{\mathrm{c}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(\frac{1}{\sqrt{3}}\right)}^{2}+{\left(\frac{1}{\sqrt{3}}\right)}^{2}+{\left(\frac{1}{\sqrt{3}}\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{\frac{3}{3}}=1\end{array}$

Q.2 Write two different vectors having same magnitude.

Ans

$\begin{array}{l}\begin{array}{l}\text{Two different vectors are:}\end{array}\\ \begin{array}{l}\stackrel{\to }{\mathrm{a}}\text{ }=\stackrel{^}{\mathrm{i}}\text{ }+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}};\text{ }\stackrel{\to }{\mathrm{b}}\text{ }=\text{ }2\stackrel{^}{\mathrm{i}}\text{ }+3\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\end{array}\\ \begin{array}{l}\text{Then}\end{array}\\ \text{Magnitude of }\stackrel{\to }{\mathrm{a}}\text{ = }\left|\stackrel{^}{\mathrm{i}}\text{ }+\text{ }2\stackrel{^}{\mathrm{j}}+\text{ }3\stackrel{^}{\mathrm{k}}\text{ }\right|\\ \begin{array}{l}\text{= }\sqrt{{\text{1}}^{\text{2}}{\text{+2}}^{\text{2}}{\text{+3}}^{\text{2}}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{1+4+9}}\end{array}\\ \begin{array}{l}\text{= }\sqrt{\text{14}}\end{array}\\ \begin{array}{l}\text{Magnitude of }\stackrel{\to }{\mathrm{b}}=\left|2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}\text{=}\sqrt{{\text{2}}^{\text{2}}{\text{+3}}^{\text{2}}{\text{+1}}^{\text{2}}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{4+9+1}}\end{array}\\ \begin{array}{l}\text{=}\sqrt{\text{14}}\end{array}\\ \begin{array}{l}\text{Hence,}\stackrel{\to }{\mathrm{a}}\text{ and }\stackrel{\to }{\mathrm{b}}\text{ are two different vectors having the same}\end{array}\\ \begin{array}{l}\text{magnitude.}\end{array}\\ \begin{array}{l}\text{The vectors are different because they have different directions.}\end{array}\end{array}$

Q.3 Write two different vectors having same direction.

Ans

$\begin{array}{l}\begin{array}{l}\text{Two different vectors are:}\end{array}\\ \begin{array}{l}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}\text{ }+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}};\text{ }\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}\text{ }+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\end{array}\\ \begin{array}{l}\text{Then,}\end{array}\\ \begin{array}{l}\text{Magnitude of }\stackrel{\to }{\text{a}}\end{array}\text{ }=\text{ }\begin{array}{l}\left|2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}=\text{ }\sqrt{{2}^{2}+{2}^{2}+{2}^{2}}\end{array}\\ \begin{array}{l}=\text{ }\sqrt{4+4+4}\end{array}\\ \begin{array}{l}=\text{ }\sqrt{12}\end{array}\\ =\text{ }\begin{array}{l}2\sqrt{3}\end{array}\\ \text{Direction cosines of }\stackrel{\to }{\mathrm{a}}\text{ }\mathrm{are}:\\ \begin{array}{l}\frac{2}{2\sqrt{3}},\frac{2}{2\sqrt{3}},\frac{2}{2\sqrt{3}}\text{ }\mathrm{i}.\mathrm{e}.,\text{ }\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\end{array}\\ \begin{array}{l}\text{Magnitude of}\stackrel{\to }{\mathrm{b}}=\left|\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right|\end{array}\\ \begin{array}{l}=\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\end{array}\\ \begin{array}{l}=\sqrt{1+1+1}\end{array}\\ \begin{array}{l}=\sqrt{3}\end{array}\\ \begin{array}{l}\text{Direction cosines of }\stackrel{\to }{\mathrm{b}}\text{ are:}\end{array}\\ \begin{array}{l}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\end{array}\\ \begin{array}{l}\text{Direction cosines of }\stackrel{\to }{\mathrm{a}}\text{ }\mathrm{and}\text{ }\stackrel{\to }{\mathrm{b}}\text{ are same. Hence, these two}\end{array}\\ \begin{array}{l}\text{vectors have same direction.}\end{array}\end{array}$

Q.4

$\begin{array}{l}\text{Find the values of}\mathrm{x}\text{and}\mathrm{y}\text{so that the vectors\hspace{0.17em}\hspace{0.17em}}2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}\text{and}\\ \mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}\text{are equal.}\end{array}$

Ans

$\begin{array}{l}\text{Since,}\\ \text{\hspace{0.17em}\hspace{0.17em}}2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}\\ \text{So, on comparing coefficients of}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}and\hspace{0.17em}}\stackrel{^}{\mathrm{j}},\text{we get}\\ \mathrm{x}=2\text{and}\mathrm{y}=3.\end{array}$

Q.5 Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).

Ans

$\begin{array}{l}\text{The vector with the initial point P (2, 1) and terminal point}\\ \text{Q (-5, 7) can be given by,}\\ \stackrel{\to }{\mathrm{PQ}}=\left(-5-2\right)\stackrel{^}{\mathrm{i}}+\left(7-1\right)\stackrel{^}{\mathrm{j}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}\\ \text{Hence, the required scalar components are -7 and 6 while the}\\ \text{vector components are\hspace{0.17em}\hspace{0.17em}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}6\stackrel{^}{\mathrm{j}}.\end{array}$

Q.6

$\begin{array}{l}\text{Find the sum of the vectors}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=-2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\\ \text{and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}.\end{array}$

Ans

$\begin{array}{l}\text{The given vectors are:}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=-2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}\\ \text{Then,}\\ \stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}+\left(-2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\right)+\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{j}}-7\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ =-\text{\hspace{0.17em}}4\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\end{array}$

Q.7

$\begin{array}{l}\text{Find the unit vector in the direction of the vector\hspace{0.17em}}\\ \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\text{The unit vector}\stackrel{^}{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}in the direction of vector\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}\hspace{0.17em}is given}\\ \text{by}\stackrel{^}{\mathrm{a}}=\frac{\stackrel{\to }{\mathrm{a}}}{|\stackrel{\to }{\mathrm{a}}|}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{{1}^{2}+{1}^{2}+{2}^{2}}}\\ =\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{6}}\\ =\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{6}}+\frac{\stackrel{^}{\mathrm{j}}}{\sqrt{6}}+\frac{2}{\sqrt{6}}\stackrel{^}{\mathrm{k}}\end{array}$

Q.8

$\begin{array}{l}\text{Find the unit vector in the direction of the vector\hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}},\text{}\\ \text{where P and Q are the points}\left(\text{1,2,3}\right)\text{and}\left(\text{4,5,6}\right)\\ \text{respectively.}\end{array}$

Ans

$\begin{array}{l}\text{Here, }\stackrel{\to }{\text{OP}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{OQ}}=4\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\\ \text{So, \hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}}=\stackrel{\to }{\mathrm{OQ}}-\stackrel{\to }{\mathrm{OP}}\\ =\left(4\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)-\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \text{Unit vector of \hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{PQ}}}{\left|\stackrel{\to }{\mathrm{PQ}}\right|}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{\left|3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right|}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{\sqrt{{3}^{2}+{3}^{2}+{3}^{2}}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{3\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}}{3\sqrt{3}}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}}{\sqrt{3}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{3}}+\frac{\stackrel{^}{\mathrm{j}}}{\sqrt{3}}+\frac{\stackrel{^}{\mathrm{k}}}{\sqrt{3}}\end{array}$

Q.9

$\begin{array}{l}\text{For given vectors,\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=-\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}},\text{find the}\\ \text{unit vector in the direction of the vector}\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}.\end{array}$

Ans

$\begin{array}{l}\text{The\hspace{0.17em}\hspace{0.17em}given vectors are:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}=-\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\left(-\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}\\ |\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}|=|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}|\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{1}^{2}+{1}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{2}\\ \therefore \text{Unit vector of \hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}}{|\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}}+\stackrel{\to }{\mathrm{b}}|}=\frac{\stackrel{^}{\mathrm{i}}}{\sqrt{2}}+\frac{\stackrel{^}{\mathrm{k}}}{\sqrt{2}}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Find}\mathrm{a}\mathrm{vector}\mathrm{in}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{vector}5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\mathrm{which}\mathrm{has}\\ \mathrm{magnitude}8\mathrm{units}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{unit}\mathrm{vector}\mathrm{of}5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}=\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{|5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}|}\\ =\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{{5}^{2}+{\left(-1\right)}^{2}+{2}^{2}}}\\ =\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{25+1+4}}\\ =\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{30}}\\ \mathrm{Thus},\mathrm{the}\mathrm{vector}\mathrm{in}\mathrm{the}\mathrm{direction}\mathrm{of}\left(5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\mathrm{which}\mathrm{has}\\ \mathrm{magnitude}8\mathrm{units}\mathrm{is}:\\ 8\stackrel{^}{\mathrm{a}}=8×\frac{5\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{\sqrt{30}}\\ =\frac{40\mathrm{ }\stackrel{^}{\mathrm{i}}-8\stackrel{^}{\mathrm{j}}+16\mathrm{ }\stackrel{^}{\mathrm{k}}}{\sqrt{30}}\\ =\frac{40}{\sqrt{30}}\mathrm{ }\stackrel{^}{\mathrm{i}}-\frac{8}{\sqrt{30}}\stackrel{^}{\mathrm{j}}+\frac{16}{\sqrt{30}}\mathrm{ }\stackrel{^}{\mathrm{k}}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{vectors}\mathrm{}2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\mathrm{and}-4\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}-8\stackrel{^}{\mathrm{k}} \mathrm{are}\\ \mathrm{collinear}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{vectors}\mathrm{are}\\ \stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{b}}=-4\stackrel{^}{\mathrm{i}}+6\stackrel{^}{\mathrm{j}}–8\stackrel{^}{\mathrm{k}}\\ =-2\left(2\stackrel{^}{\mathrm{i}}–3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)\\ \stackrel{\to }{\mathrm{b}}\mathrm{ }=-2\stackrel{\to }{\mathrm{a}},\mathrm{which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\stackrel{\to }{\mathrm{a}}=\mathrm{\lambda }\stackrel{\to }{\mathrm{b}},\\ \mathrm{where},\mathrm{\lambda }=-2\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{vectors}\mathrm{are}\mathrm{collinear}.\end{array}$

Q.12

$\text{Find the direction cosines of the vector}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}.$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{vector} \mathrm{is}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}.\\ |\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}|=\sqrt{{1}^{2}+{2}^{2}+{3}^{2}}\\ =\sqrt{1+4+9}\\ =\sqrt{14}\\ \mathrm{So},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}} \mathrm{are} \left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right).\end{array}$

Q.13 Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, – 2, 1) directed A to B.

Ans

$\begin{array}{l}\text{The given points are}\mathrm{A}\left(1,\text{}2,–3\right)\text{and}\mathrm{B}\left(–1,–2,1\right).\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}=\left(-1-1\right)\stackrel{^}{\mathrm{i}}+\left(-2-2\right)\stackrel{^}{\mathrm{j}}+\left(1+3\right)\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-2\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \therefore |\stackrel{\to }{\mathrm{AB}}|=\sqrt{{\left(-2\right)}^{2}+{\left(-4\right)}^{2}+{4}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{4+16+16}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{36}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ \text{Hence, the direction cosines of \hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}\hspace{0.17em}are\hspace{0.17em}\hspace{0.17em}}\left(\frac{-2}{6},\frac{-4}{6},\frac{4}{6}\right)=\left(\frac{-1}{3},\frac{-2}{3},\frac{2}{3}\right).\end{array}$

Q.14

$\begin{array}{l}\text{Show that the vector}\stackrel{^}{\text{i}}+\stackrel{^}{\text{j}}+\stackrel{^}{\text{k}}\text{is equally inclined to the axes}\\ \text{OX, OY and OZ.}\end{array}$

Ans

$\begin{array}{l}\text{The given vector is}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\stackrel{\to }{\mathrm{a}}|=|\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\\ =\sqrt{3}\\ \text{The direction cosines of}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}are}\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right).\\ \text{Let}\mathrm{\alpha },\text{\hspace{0.17em}}\mathrm{\beta }\text{and}\mathrm{\gamma }\text{be the angles formed by}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}with the positive}\\ \text{directions of x, y, and z axes.}\\ \text{Then, we have cos}\mathrm{\alpha }=\frac{1}{\sqrt{3}},\text{cos}\mathrm{\beta }=\frac{1}{\sqrt{3}}\text{,\hspace{0.17em}cos}\mathrm{\gamma }=\frac{1}{\sqrt{3}}.\\ \text{Hence, the given vector is equally inclined to axes OX,}\\ \text{OY, and OZ.}\end{array}$

Q.15

$\begin{array}{l}\text{Find the position vector of a point R which divides the}\\ \text{line joining two points P and Q whose position vectors}\\ \text{are\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}–\stackrel{^}{\mathrm{k}}\text{and}–\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{\hspace{0.17em}}\mathrm{k}}\text{respectively, in the ration 2:1}\\ \left(\text{i}\right)\text{internally\hspace{0.17em}}\\ \left(\text{ii}\right)\text{\hspace{0.17em}externally}\end{array}$

Ans

$\begin{array}{l}\text{The given vectors are}\stackrel{\to }{\mathrm{OP}}=\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{OQ}}=-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{\hspace{0.17em}}\mathrm{k}}.\\ \text{The position vector of point R dividing the line segment joining}\\ \text{two points\hspace{0.17em}P and Q in the ratio 2: 1 is given by:}\\ \left(\text{i}\right)\text{Internally:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OR}}=\frac{2\left(-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{\hspace{0.17em}}\mathrm{k}}\right)+1\left(\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)}{2+1}\\ =\frac{-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+4\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}}{3}\\ =-\frac{1}{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\frac{4}{3}\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+\frac{1}{3}\stackrel{^}{\mathrm{k}}\\ \left(\text{ii}\right)\text{The position vector of point R which divides the line joining}\\ \text{two points P and Q externally in the ratio 2:1 is given by,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OR}}=\frac{2\left(-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\text{\hspace{0.17em}}\mathrm{k}}\right)-1\left(\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)}{2-1}\\ =\frac{-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+0\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}+3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}}{1}\\ =-3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+3\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\end{array}$

Q.16 Find the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).

Ans

$\begin{array}{l}\text{The position vector of mid-point R of the vector joining points}\\ \text{P (2, 3, 4) and Q (4, 1,-2) is given by,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{OR}}=\frac{\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\left(4\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)}{2}\\ =\frac{6\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}}{2}\\ =3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\end{array}$

Q.17

$\begin{array}{l}\text{Show that the points A, B and C with position vectors,}\\ \stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}},\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\text{respectively}\\ \text{form the vertices of a right angled triangle.}\end{array}$

Ans

$\begin{array}{l}\text{Position vectors of points A, B, and C are respectively given as:}\\ \stackrel{\to }{\mathrm{a}}=3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}and}\stackrel{\to }{\mathrm{c}}=\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{b}}-\stackrel{\to }{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)-\left(3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}=\stackrel{\to }{\mathrm{c}}-\stackrel{\to }{\mathrm{b}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)-\left(2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{\mathrm{a}}-\stackrel{\to }{\mathrm{c}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\right)-\left(\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)\end{array}$ $\begin{array}{c}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \text{So,\hspace{0.17em}}{|\stackrel{\to }{\mathrm{AB}}|}^{2}={|-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=35\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{BC}}|}^{2}={|-\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-6\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=41\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{CA}}|}^{2}={|2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}|}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ {|\stackrel{\to }{\mathrm{AB}}|}^{2}=\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{|\stackrel{\to }{\mathrm{BC}}|}^{2}+{|\stackrel{\to }{\mathrm{CA}}|}^{2}\\ =36+5=41\end{array}$ $\text{Hence, ABC is a right-angled triangle}$

Q.18

$\mathrm{In}\mathrm{triangle},\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{not}\mathrm{true}:\phantom{\rule{0ex}{0ex}}\begin{array}{l}\left(\mathrm{A}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=0\\ \left(\mathrm{B}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}–\mathrm{AC}=0\\ \left(\mathrm{C}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{BC}}–\stackrel{\to }{\mathrm{CA}}=0\\ \left(\mathrm{D}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CB}}+\stackrel{\to }{\mathrm{CA}}=0\end{array}$

Ans

$\begin{array}{l}\text{On applying the triangle law of addition in the given triangle,}\\ \text{we have:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=\stackrel{\to }{\mathrm{AC}}...\left(\text{i}\right)\\ ⇒\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=-\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}\\ ⇒\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{0}...\left(\text{ii}\right)\\ \therefore \text{The equation given in alternative A is true.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}\\ ⇒\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}-\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0}\\ \therefore \text{The equation given in alternative B is true.}\\ \text{From equation}\left(\text{ii}\right)\text{, we have:}\\ \stackrel{\to }{\mathrm{AB}}-\stackrel{\to }{\mathrm{CB}}+\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0}\\ \therefore \text{The equation given in alternative D is true.}\\ \text{Now, the equation given in alternative C,}\\ \stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}-\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=\stackrel{\to }{\mathrm{CA}}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{From equation}\left(\text{i}\right)\text{and}\left(\text{iii}\right)\text{,​\hspace{0.17em}we have}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{CA}}=\stackrel{\to }{\mathrm{AC}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{\mathrm{AC}}\\ \text{\hspace{0.17em}}⇒\stackrel{\to }{\mathrm{AC}}+\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{0},\text{which is not true.}\\ \text{Hence, the equation given in alternative C is incorrect.}\\ \text{The correct answer is C.}\end{array}$

Q.19

$\begin{array}{l}\text{If}\stackrel{\to }{\mathrm{a}}\text{}\stackrel{\to }{\mathrm{b}}\text{are two collinear vectors, then which of the following}\\ \text{are incorrect:}\\ \left(\text{A}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\mathrm{\lambda }\stackrel{\to }{\mathrm{a}},\text{for some scalar}\mathrm{\lambda }\\ \left(\text{B}\right)\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}=±\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}\\ \left(\text{C}\right)\text{the respective components of}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{are proportional}\\ \left(\text{D}\right)\text{both the vectors}\stackrel{\to }{\mathrm{a}}\text{and}\stackrel{\to }{\mathrm{b}}\text{have same direction, but}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}different magnitudes.}\end{array}$

Ans

$\begin{array}{l}\text{If}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{are two collinear vectors, then they are parallel.}\\ \text{Therefore, we have:}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\mathrm{\lambda }\stackrel{\to }{\mathrm{a}}\left(\mathrm{\lambda }\text{is a scalar quantities.}\right)\\ \text{If\hspace{0.17em}}\mathrm{\lambda }=±1,\text{then \hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=±1\stackrel{\to }{\mathrm{a}}.\\ \text{If\hspace{0.17em}}\stackrel{\to }{\mathrm{a}}={\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{b}}={\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}then}\\ \text{\hspace{0.17em}\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\mathrm{\lambda }\stackrel{\to }{\mathrm{a}}⇒{\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}=\mathrm{\lambda }\left({\mathrm{a}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\right)\end{array}$ $\begin{array}{l}⇒{\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}=\left({\mathrm{\lambda a}}_{1}\right)\stackrel{^}{\mathrm{i}}+\left({\mathrm{\lambda a}}_{2}\right)\stackrel{^}{\mathrm{j}}+\left({\mathrm{\lambda a}}_{3}\right)\stackrel{^}{\mathrm{k}}\\ ⇒{\mathrm{b}}_{1}={\mathrm{\lambda a}}_{1},{\mathrm{b}}_{2}={\mathrm{\lambda a}}_{2},{\mathrm{b}}_{3}={\mathrm{\lambda a}}_{3}\\ ⇒\frac{{\mathrm{b}}_{1}}{{\mathrm{a}}_{1}}=\frac{{\mathrm{b}}_{2}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{3}}{{\mathrm{a}}_{3}}=\mathrm{\lambda }\\ \text{Thus, the respective components of}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}are proportional.}\\ \text{However, vectors}\stackrel{\to }{\mathrm{a}}\text{\hspace{0.17em}\hspace{0.17em}and}\stackrel{\to }{\mathrm{b}}\text{\hspace{0.17em}\hspace{0.17em}can have different directions.}\\ \text{Hence, the statement given in D is incorrect.}\\ \text{The correct answer is D.}\end{array}$

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### 1. How important is it to practice all the questions in Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2?

Mathematics is a subject that requires ample amount of practice and for scoring good in the Mathematics Class 12 board examinations and having clarity of the concepts, the students must practice every question of the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2.

### 2. Are the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 difficult?

No, the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 are not difficult. Practising these solutions thoroughly will help the students to gain good marks in Class 12 Mathematics board examinations.

### 3. Is the NCERT Textbook enough to prepare the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2?

Yes, the NCERT Textbook is enough to strengthen the basic concepts of the students, so that they can solve any problem related to these concepts. Students should also practice sample papers and the past years’ papers for better preparation. Practising the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 help students to have a deep understanding of the concepts and eventually leads to better scores.

### 4. How can the students clear their doubts about the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2?

Students can refer to the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 provided by the Extramarks’ website to clear their doubts. Also, students can subscribe to the website and have access to learning tools such as live doubt-solving classes, self-assessment, in-depth performance reports and much more. Students can also record their lectures for further help.

### 5. Are the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 available on Extramarks?

Yes, the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 are available on the Extramarks’ website. They are easily accessible as they are in PDF Format, and can also be downloaded on any device.

### 6. How many books are there in the curriculum of Class 12 Mathematics?

There are two volumes of NCERT Textbooks in the curriculum of Class 12 Mathematics. These books in themselves are enough for building the fundamentals of the students. This helps students to have strong concepts of Mathematics, ultimately leading to better scores.

### 7. Does the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 assist students in other fields of study despite the board examinations?

The students who want to opt for Engineering in their higher education should be very keen with the concepts of Mathematics. The entrance examinations of such as JEE include the concepts of NCERT as it forms the core of the subject. Also, the Common Entrance Test of Delhi University is entirely based on the NCERT curriculum. Therefore, the Class 12 Maths NCERT Solutions Chapter 10 Exercise 10.2 assist students in various competitive examinations.

### 8. Does practising the NCERT Exemplar questions help students score well in Class 12 Mathematics board examinations?

Yes, students should thoroughly practice every question that appears in front of them. Every question helps them understand the concepts more deeply and helps them perform effectively in the board examinations. Therefore, students should review the NCERT Exemplar questions thoroughly before any examination.