Vector Algebra

Class 12 NCERT

NCERT

1   Represent graphically a displacement of $ 40 \ km, 30^o$ east of north.

Solution :

Here, vector $OP$ represents the displacement of $ 40 km, 30^o$ East of North.

2   Classify the following measures as scalars and vectors.$\\$ $(i) 10 kg $$\\$ $(ii) 2$ meters north-west$\\$$ (iii) 40^o $$\\$$(iv) 40 $ watt$\\$ $(v) 10^{-19}$ coulomb $\\$$(vi) 20 \ m / s ^2$

Solution :

$(i) 10 \ kg $ is a scalar quantity because it involves only magnitude.$\\$ $(ii) 2 $ meters north-west is a vector quantity as it involves both magnitude and direction.$\\$ $(iii) 40^o $ is a scalar quantity as it involves only magnitude.$\\$ $(iv) 40$ watts is a scalar quantity as it involves only magnitude.$\\$ $(v) 10^{-19}$ Coulomb is a scalar quantity as it involves only magnitude.$\\$ $(vi) 20 m / s^2$ is a vector quantity as it involves magnitude as well as direction.

3   Classify the following as scalar and vector quantities.$\\$ (i) time period $\\$(ii) distance$\\$ (iii) force$\\$ (iv) velocity $\\$(v) work done

Solution :

(i) Time period is a scalar quantity as it involves only magnitude. $\\$(ii) Distance is a scalar quantity as it involves only magnitude. $\\$(iii) Force is a vector quantity as it involves both magnitude and direction.$\\$ (iv) Velocity is a vector quantity as it involves both magnitude as well as direction.$\\$ (v) Work done is a scalar quantity as it involves only magnitude.

4   In Figure, identify the following vectors.$\\$ (i) Co-initial$\\$ (ii) Equal$\\$ (iii) Collinear but not equal

Solution :

(i) Vectors $a $ and $d$ are co-initial because they have the same initial point.$\\$ (ii) Vectors $b$ and $d$ are equal because they have the same magnitude and direction.$\\$ (iii) Vectors $a$ and $c$ are collinear but not equal. This is because although they are parallel, their directions are not the same

5   Answer the following as true or false:$\\$ (i) $\overrightarrow{a}$ and $- \overrightarrow{a}$ are collinear.$\\$ (ii) Two collinear vectors are always equal in magnitude.$\\$ (iii)Two vectors having same magnitude are collinear.$\\$ (iv) Two collinear vectors having the same magnitude are equal.

Solution :

(i) True$\\$ $\ \ \ \ \ \ \ $Vectors $\overrightarrow{a}$ and $-\overrightarrow{ a}$ can be parallel or coinciding vectors. Either way the vectors will have same magnitude but opposite in direction and will be parallel to the same line..$\\$ (ii) False$\\$ $\ \ \ \ \ \ \ $Collinear vectors are those vectors that are parallel to the same line.$\\$ (iii) False$\\$ $\ \ \ \ \ \ \ $It is not necessary for two vectors having the same magnitude to be parallel to the same line.$\\$ (iv) False$\\$ $\ \ \ \ \ \ \ $Two vectors are said to be equal if they have the same magnitude and direction, regardless of the positions of their initial points.

6   Compute the magnitude of the following vectors:$\\$ $\vec{a}=\hat{i}+\hat{j}+\hat{k}; \ \vec{b}=2\hat{i}-7\hat{j}-3\hat{k}; \ \vec{c}=\dfrac{1}{\sqrt{3}}\hat{i}+\dfrac{1}{\sqrt{3}}\hat{j}-\dfrac{1}{\sqrt{3}}\hat{k}$

Solution :

$\vec{a}=\hat{i}+\hat{j}+\hat{k}; \ \vec{b}=2\hat{i}-7\hat{j}-3\hat{k}; \ \vec{c}=\dfrac{1}{\sqrt{3}}\hat{i}+\dfrac{1}{\sqrt{3}}\hat{j}-\dfrac{1}{\sqrt{3}}\hat{k}$$\\$ Magnitude of a vector $\vec{v}=p\hat{i}+q\hat{j}+r\hat{k}$ is given by $|\vec{v}|=\sqrt{(p)^2+(q)^2+(r)^2}$$\\$ $|\vec{a}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}\\ |\vec{b}|=\sqrt{(2)^2+(-7)^2+(-3)^2}\\ =\sqrt{4+49+9}\\ =\sqrt{62}\\ |\vec{c}|=\sqrt{(\dfrac{1}{\sqrt{3}})^2+(\dfrac{1}{\sqrt{3}})^2+(-\dfrac{1}{\sqrt{3}})^2}\\ =\sqrt{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}}=1$

7   Write two different vectors having same magnitude.

Solution :

Consider $\vec{a}=(\hat{i}+2\hat{j}+3 \hat{k})$ and $ \vec{b}=(2\hat{i}-\hat{j}-3 \hat{k})$$\\$ It can be observed that $|\vec{a}|=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$ and $ \\$ $|\vec{b}|=\sqrt{2^2+(-1)^2+(-3)^2}=\sqrt{4+1+9}=\sqrt{14}$$\\$ Hence, $\vec{a}$ and $\vec{b}$ are two different vectors having the same magnitude. The vectors are different because they have different directions.

8   Write two different vectors having same direction.

Solution :

Consider $\vec{p}=(\hat{i}+\hat{j}+\hat{k})$ and $ \vec{q}=(2 \hat{i}+2\hat{j}+2\hat{k})$$\\$ The direction cosines of $\vec{p}$ are given by,$\\$ $l=\dfrac{1}{\sqrt{1^2+1^2+1^2}}=\dfrac{1}{\sqrt{3}},\\ m=\dfrac{1}{\sqrt{1^2+1^2+1^2}}=\dfrac{1}{\sqrt{3}},\\ n=\dfrac{1}{\sqrt{1^2+1^2+1^2}}=\dfrac{1}{\sqrt{3}}$$\\$ The direction cosines of $\vec{q}$ are given by$\\$ $l=\dfrac{2}{\sqrt{2^2+2^2+2^2}}=\dfrac{2}{2\sqrt{3}}=\dfrac{1}{\sqrt{3}},\\ m=\dfrac{2}{\sqrt{2^2+2^2+2^2}}=\dfrac{2}{2\sqrt{3}}=\dfrac{1}{\sqrt{3}},$$\\$ and $ n=\dfrac{2}{\sqrt{2^2+2^2+2^2}}=\dfrac{2}{2\sqrt{3}}=\dfrac{1}{\sqrt{3}}$$\\$ The direction cosines of $\vec{p}$ and $\vec{q}$ are the same. Hence, the two vectors have the same direction.

9   Find the values of x and y so that the vectors $2\hat{i}+3\hat{j}$ and $ x\hat{i}+y\hat{j}$ are equal$\\$

Solution :

The two vectors $2\hat{i}+3\hat{j}$ and $ x\hat{i}+y\hat{j}$ will be equal if their corresponding components are equal. Hence, the required values of $x$ and $y$ are $2$ and $3 $respectively.

10   Find the scalar and vector components of the vector with initial point $(2,1)$ and terminal point $(-5,7)$

Solution :

The vector with the initial point $P(2,1)$and terminal point $Q(-5,7)$ can be given by,$\\$ $\vec{PQ}=(-5-2)\vec{i}+(7-1)\vec{j}\\ \Rightarrow \vec{PQ}=-7\hat{i}+6\hat{j}$$\\$ Hence, the required scalar components are $-7$ and $6$ while the vector components are $-7\hat{i}+6\vec{j}$