Solution

It is given that,

$|\vec{a}|=\sqrt{3},|\vec{b}|=2$ and, $\vec{a} \cdot \vec{b}=\sqrt{6}$

Now, we know that $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$.

$\therefore \sqrt{6}=\sqrt{3} \times 2 \times \cos \theta$

$\Rightarrow \cos \theta=\frac{\sqrt{6}}{\sqrt{3} \times 2}$

$\Rightarrow \cos \theta=\frac{1}{\sqrt{2}}$

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

Hence, the angle between the given vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$.

Solution

The given vectors are $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}$.

$ \begin{aligned} & |\vec{a}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14} \\ & |\vec{b}|=\sqrt{3^{2}+(-2)^{2}+1^{2}}=\sqrt{9+4+1}=\sqrt{14} \end{aligned} $

Now, $\vec{a} \cdot \vec{b}=(\hat{i}-2 \hat{j}+3 \hat{k})(3 \hat{i}-2 \hat{j}+\hat{k})$

$ \begin{aligned} & =1.3+(-2)(-2)+3.1 \\ & =3+4+3 \\ & =10 \end{aligned} $

Also, we know that $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$

$\therefore 10=\sqrt{14} \sqrt{14} \cos \theta$

$\Rightarrow \cos \theta=\frac{10}{14}$

$\Rightarrow \theta=\cos ^{-1}(\frac{5}{7})$

Solution

Let $\vec{a}=\hat{i}-\hat{j}$ and $\vec{b}=\hat{i}+\hat{j}$.

Now, projection of vector $\vec{a}$ on $\vec{b}$ is given by,

$\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{1}{\sqrt{1+1}}{1.1+(-1)(1)}=\frac{1}{\sqrt{2}}(1-1)=0$

Hence, the projection of vector $\vec{a}$ on $\vec{b}$ is 0 .

Solution

Let $\vec{a}=\hat{i}+3 \hat{j}+7 \hat{k}$ and $\hat{b}=7 \hat{i}-\hat{j}+8 \hat{k}$.

Now, projection of vector $\vec{a}$ on $\vec{b}$ is given by, $\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{1}{\sqrt{7^{2}+(-1)^{2}+8^{2}}}{1(7)+3(-1)+7(8)}=\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}$

Solution

Let $\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})=\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}$,

$\vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}$,

$\vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})=\frac{6}{7} \hat{i}+\frac{2}{7} \hat{j}-\frac{3}{7} \hat{k}$.

$|\vec{a}|=\sqrt{(\frac{2}{7})^{2}+(\frac{3}{7})^{2}+(\frac{6}{7})^{2}}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$

$|\vec{b}|=\sqrt{(\frac{3}{7})^{2}+(-\frac{6}{7})^{2}+(\frac{2}{7})^{2}}=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}}=1$

$|\vec{c}|=\sqrt{(\frac{6}{7})^{2}+(\frac{2}{7})^{2}+(-\frac{3}{7})^{2}}=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=1$

Thus, each of the given three vectors is a unit vector.

$\vec{a} \cdot \vec{b}=\frac{2}{7} \times \frac{3}{7}+\frac{3}{7} \times(\frac{-6}{7})+\frac{6}{7} \times \frac{2}{7}=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=0$

$\vec{b} \cdot \vec{c}=\frac{3}{7} \times \frac{6}{7}+(\frac{-6}{7}) \times \frac{2}{7}+\frac{2}{7} \times(\frac{-3}{7})=\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=0$

$\vec{c} \cdot \vec{a}=\frac{6}{7} \times \frac{2}{7}+\frac{2}{7} \times \frac{3}{7}+(\frac{-3}{7}) \times \frac{6}{7}=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0$

Hence, the given three vectors are mutually perpendicular to each other.

Solution

$(\vec{a} \cdot \vec{b}) \cdot(\vec{a}-\vec{b})=8$

$\Rightarrow \vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}=8$

$\Rightarrow|\vec{a}|^{2}-|\vec{b}|^{2}=8$

$\Rightarrow(8|\vec{b}|)^{2}-|\vec{b}|^{2}=8 \quad[|\vec{a}|=8|\vec{b}|]$

$\Rightarrow 64|\vec{b}|^{2}-|\vec{b}|^{2}=8$

$\Rightarrow 63|\vec{b}|^{2}=8$

$\Rightarrow|\vec{b}|^{2}=\frac{8}{63}$

$\Rightarrow|\vec{b}|=\sqrt{\frac{8}{63}} \quad$ [Magnitude of a vector is non-negative]

$\Rightarrow|\vec{b}|=\frac{2 \sqrt{2}}{3 \sqrt{7}}$

$|\vec{a}|=8|\vec{b}|=\frac{8 \times 2 \sqrt{2}}{3 \sqrt{7}}=\frac{16 \sqrt{2}}{3 \sqrt{7}}$

Solution

$ \begin{aligned} & (3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b}) \\ & =3 \vec{a} \cdot 2 \vec{a}+3 \vec{a} \cdot 7 \vec{b}-5 \vec{b} \cdot 2 \vec{a}-5 \vec{b} \cdot 7 \vec{b} \\ & =6 \vec{a} \cdot \vec{a}+21 \vec{a} \cdot \vec{b}-10 \vec{a} \cdot \vec{b}-35 \vec{b} \cdot \vec{b} \\ & =6|\vec{a}|^{2}+11 \vec{a} \cdot \vec{b}-35|\vec{b}|^{2} \end{aligned} $

Solution

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$.

It is given that $|\vec{a}|=|\vec{b}|, \vec{a} \cdot \vec{b}=\frac{1}{2}$, and $\theta=60^{\circ}$.

We know that $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$

$\therefore \frac{1}{2}=|\vec{a}||\vec{a}| \cos 60^{\circ}$

[Using (1)]

$\Rightarrow \frac{1}{2}=|\vec{a}|^{2} \times \frac{1}{2}$

$\Rightarrow|\vec{a}|^{2}=1$

$\Rightarrow|\vec{a}|=|\vec{b}|=1$

Solution

$(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$

$\Rightarrow \vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\bar{{}a} \cdot \vec{a}=12$

$\Rightarrow|\vec{x}|^{2}-|\vec{a}|^{2}=12$

$\Rightarrow|\vec{x}|^{2}-1=12 \quad[|\vec{a}|=1$ as $\vec{a}$ is a unit vector $]$

$\Rightarrow|\vec{x}|^{2}=13$

$\therefore|\vec{x}|=\sqrt{13}$

Solution

The given vectors are $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$, and $\vec{c}=3 \hat{i}+\hat{j}$.

Now,

$\vec{a}+\lambda \vec{b}=(2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})=(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}$

If $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$, then

$(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0$.

$\Rightarrow[(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}] \cdot(3 \hat{i}+\hat{j})=0$

$\Rightarrow(2-\lambda) 3+(2+2 \lambda) 1+(3+\lambda) 0=0$

$\Rightarrow 6-3 \lambda+2+2 \lambda=0$

$\Rightarrow-\lambda+8=0$

$\Rightarrow \lambda=8$

Hence, the required value of $\lambda$ is 8 .

Solution

$(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}) \cdot(|\vec{a}| \vec{b}-|\vec{b}| \vec{a})$

$=|\vec{a}|^{2} \vec{b} \cdot \vec{b}-|\vec{a}||\vec{b}| \vec{b} \cdot \vec{a}+|\vec{b}||\vec{a}| \vec{a} \cdot \vec{b}-|\vec{b}|^{2} \vec{a} \cdot \vec{a}$

$=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{b}|^{2}|\vec{a}|^{2}$

$=0$

Hence, $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ and $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$ are perpendicular to each other.

Solution

It is given that $\vec{a} \cdot \vec{a}=0$ and $\vec{a} \cdot \vec{b}=0$.

Now,

$\vec{a} \cdot \vec{a}=0 \Rightarrow|\vec{a}|^{2}=0 \Rightarrow|\vec{a}|=0$

$\therefore \vec{a}$ is a zero vector.

Hence, vector $\vec{b}$ satisfying $\vec{a} \cdot \vec{b}=0$ can be any vector.

Solution

We have $|\overrightarrow{\mathrm{a}}|=1,|\overrightarrow{\mathrm{b}}|=1,|\overrightarrow{\mathrm{c}}|=1$

Also $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$

Squaring we get, $$ \begin{aligned} & |\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2+|\overrightarrow{\mathrm{c}}|^2+2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})=0 \\ & \Rightarrow 1+1+1+2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})=0 \\ & \therefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=-\frac{3}{2} \end{aligned} $$

Solution

Consider $\vec{a}=2 \hat{i}+4 \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+3 \hat{j}-6 \hat{k}$.

Then,

$\vec{a} \cdot \vec{b}=2.3+4.3+3(-6)=6+12-18=0$

We now observe that:

$|\vec{a}|=\sqrt{2^{2}+4^{2}+3^{2}}=\sqrt{29}$

$\therefore \vec{a} \neq \overrightarrow{{}0}$

$|\vec{b}|=\sqrt{3^{2}+3^{2}+(-6)^{2}}=\sqrt{54}$

$\therefore \vec{b} \neq \overrightarrow{{}0}$

Hence, the converse of the given statement need not be true.

Solution

The vertices of $\triangle A B C$ are given as $A(1,2,3), B(-1,0,0)$, and $C(0,1,2)$.

Also, it is given that $\square ABC$ is the angle between the vectors $\overrightarrow{{}BA}$ and $\overrightarrow{{}BC}$.

$ \begin{aligned} & \overrightarrow{{}BA}={1-(-1)} \hat{i}+(2-0) \hat{j}+(3-0) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k} \\ & \overrightarrow{{}BC}={0-(-1)} \hat{i}+(1-0) \hat{j}+(2-0) \hat{k}=\hat{i}+\hat{j}+2 \hat{k} \\ & \therefore \overrightarrow{{}BA} \cdot \overrightarrow{{}BC}=(2 \hat{i}+2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+\hat{j}+2 \hat{k})=2 \times 1+2 \times 1+3 \times 2=2+2+6=10 \\ & \mid \overrightarrow{{}BA}=\sqrt{2^{2}+2^{2}+3^{2}}=\sqrt{4+4+9}=\sqrt{17} \\ & \mid \overrightarrow{{}BC}=\sqrt{1+1+2^{2}}=\sqrt{6} \end{aligned} $

Now, it is known that:

$\overrightarrow{{}BA} \cdot \overrightarrow{{}BC}=|\overrightarrow{{}BA}||\overrightarrow{{}BC}| \cos (\angle ABC)$

$\therefore 10=\sqrt{17} \times \sqrt{6} \cos (\angle ABC)$

$\Rightarrow \cos (\angle ABC)=\frac{10}{\sqrt{17} \times \sqrt{6}}$

$\Rightarrow \angle ABC=\cos ^{-1}(\frac{10}{\sqrt{102}})$

Solution

The given points are $A(1,2,7), B(2,6,3)$, and $C(3,10,-1)$.

$\therefore \overrightarrow{{}AB}=(2-1) \hat{i}+(6-2) \hat{j}+(3-7) \hat{k}=\hat{i}+4 \hat{j}-4 \hat{k}$

$ \overrightarrow{{}BC}=(3-2) \hat{i}+(10-6) \hat{j}+(-1-3) \hat{k}=\hat{i}+4 \hat{j}-4 \hat{k} $

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

$|\overrightarrow{{}AB}|=\sqrt{1^{2}+4^{2}+(-4)^{2}}=\sqrt{1+16+16}=\sqrt{33}$

$|\overrightarrow{{}BC}|=\sqrt{1^{2}+4^{2}+(-4)^{2}}=\sqrt{1+16+16}=\sqrt{33}$

$|\overrightarrow{{}AC}|=\sqrt{2^{2}+8^{2}+8^{2}}=\sqrt{4+64+64}=\sqrt{132}=2 \sqrt{33}$

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

Hence, the given points A, B, and C are collinear.

Solution

Let vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$ be position vectors of points $A, B$, and $C$ respectively.

i.e., $\overrightarrow{{}OA}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{{}OB}=\hat{i}-3 \hat{j}-5 \hat{k}$ and $\overrightarrow{{}OC}=3 \hat{i}-4 \hat{j}-4 \hat{k}$

Now, vectors $\overrightarrow{{}AB}, \overrightarrow{{}BC}$, and $\overrightarrow{{}AC}$ represent the sides of $\triangle ABC$.

i.e., $\overrightarrow{{}OA}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{{}OB}=\hat{i}-3 \hat{j}-5 \hat{k}$, and $\overrightarrow{{}OC}=3 \hat{i}-4 \hat{j}-4 \hat{k}$

$\therefore \overrightarrow{{}AB}=(1-2) \hat{i}+(-3+1) \hat{j}+(-5-1) \hat{k}=-\hat{i}-2 \hat{j}-6 \hat{k}$

$\overrightarrow{{}BC}=(3-1) \hat{i}+(-4+3) \hat{j}+(-4+5) \hat{k}=2 \hat{i}-\hat{j}+\hat{k}$

$\overrightarrow{{}AC}=(2-3) \hat{i}+(-1+4) \hat{j}+(1+4) \hat{k}=-\hat{i}+3 \hat{j}+5 \hat{k}$

$|\overrightarrow{{}AB}|=\sqrt{(-1)^{2}+(-2)^{2}+(-6)^{2}}=\sqrt{1+4+36}=\sqrt{41}$

$|\overrightarrow{{}BC}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{4+1+1}=\sqrt{6}$

$|\overrightarrow{{}AC}|=\sqrt{(-1)^{2}+3^{2}+5^{2}}=\sqrt{1+9+25}=\sqrt{35}$

$\therefore|\overrightarrow{{}BC}|^{2}+|\overrightarrow{{}AC}|^{2}=6+35=41=|\overrightarrow{{}AB}|^{2}$

Hence, $\triangle A B C$ is a right-angled triangle.

Solution

Vector $\lambda \vec{a}$ is a unit vector if $|\lambda \vec{a}|=1$.

Now,

$|\lambda \vec{a}|=1$

$\Rightarrow|\lambda||\vec{a}|=1$

$\Rightarrow|\vec{a}|=\frac{1}{|\lambda|}$

$[\lambda \neq 0]$

$\Rightarrow a=\frac{1}{|\lambda|}$

$[|\vec{a}|=a]$

Hence, vector $\lambda \vec{a}$ is a unit vector if $a=\frac{1}{|\lambda|}$. The correct answer is $D$.