Solution

Given that

$ \begin{aligned} \vec{a} & =2 \hat{i}-\hat{j}+\hat{k} \text{ and } \vec{b}=2 \hat{j}+\hat{k} \\ \vec{a}+\vec{b} & =(2 \hat{i}-\hat{j}+\hat{k})+(2 \hat{j}+\hat{k})=2 \hat{i}+\hat{j}+2 \hat{k} \end{aligned} $

$\therefore \quad$ Unit vector in the direction of $\vec{a}+\vec{b}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}$

$ \begin{aligned} & =\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{\sqrt{(2)^{2}+(1)^{2}+(2)^{2}}}=\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{\sqrt{4+1+4}} \\ & =\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{\sqrt{9}}=\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{3}=\frac{2}{3} \hat{i}+\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k} \end{aligned} $

Hence, the required unit vector is $\frac{2}{3} \hat{i}+\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k}$.

Solution

Given that $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-2 \hat{k}$

(i) $6 \vec{b}=6(2 \hat{i}+\hat{j}-2 \hat{k})=12 \hat{i}+6 \hat{j}-12 \hat{k}$

$\therefore \quad$ Unit vector in the direction of $6 \vec{b}=\frac{6 \vec{b}}{|6 \vec{b}|}$

$ \begin{aligned} & =\frac{12 \hat{i}+6 \hat{j}-12 \hat{k}}{\sqrt{(12)^{2}+(6)^{2}+(-12)^{2}}}=\frac{12 \hat{i}+6 \hat{j}-12 \hat{k}}{\sqrt{144+36+144}} \\ & =\frac{12 \hat{i}+6 \hat{j}-12 \hat{k}}{\sqrt{324}}=\frac{12 \hat{i}+6 \hat{j}-12 \hat{k}}{18} \\ & =\frac{6}{18}(2 \hat{i}+\hat{j}-2 \hat{k})=\frac{1}{3}(2 \hat{i}+\hat{j}-2 \hat{k}) \end{aligned} $

Hence, the required unit vector is $\frac{1}{3}(2 \hat{i}+\hat{j}-2 \hat{k})$.

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

$ =2 \hat{i}+2 \hat{j}+4 \hat{k}-2 \hat{i}-\hat{j}+2 \hat{k}=\hat{j}+6 \hat{k} $

$\therefore \quad$ Unit vector in the direction of $2 \vec{a}-\vec{b}$

$ \begin{aligned} & =\frac{2 \vec{a}-\vec{b}}{|2 \vec{a}-\vec{b}|}=\frac{\hat{j}+6 \hat{k}}{\sqrt{(1)^{2}+(6)^{2}}}=\frac{\hat{j}+6 \hat{k}}{\sqrt{1+36}} \\ & =\frac{\hat{j}+6 \hat{k}}{\sqrt{37}}=\frac{1}{\sqrt{37}}[\hat{j}+6 \hat{k}] \end{aligned} $

Hence, the required unit vector is $\frac{1}{\sqrt{37}}[\hat{j}+6 \hat{k}]$.

Solution

Given coordinates are $P(5,0,8)$ and $Q(3,3,2)$

$\therefore \quad \overrightarrow{{}PQ}=(3-5) \hat{i}+(3-0) \hat{j}+(2-8) \hat{k}=-2 \hat{i}+3 \hat{j}-6 \hat{k}$

$\therefore \quad$ Unit vector in the direction of $\overrightarrow{{}PQ}=\frac{\overrightarrow{{}PQ}}{|\overrightarrow{{}PQ}|}$

$ \begin{aligned} & =\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{(-2)^{2}+(3)^{2}+(-6)^{2}}}=\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{4+9+36}}=\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{49}} \\ & =\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{7}=\frac{1}{7}(-2 \hat{i}+3 \hat{j}-6 \hat{k}) \end{aligned} $

Hence, the required unit vector is $\frac{1}{7}(-2 \hat{i}+3 \hat{j}-6 \hat{k})$.

Solution

Given that

$ \begin{matrix} BC =1.5 BA \\ \Rightarrow \quad \frac{BC}{BA} =1.5=\frac{3}{2} \\ \Rightarrow \quad \frac{\vec{c}-\vec{b}}{\vec{a}-\vec{b}} =\frac{3}{2} \\ \Rightarrow 2 \vec{c}-2 \vec{b} =3 \vec{a}-3 \vec{b} \Rightarrow 2 \vec{c}=3 \vec{a}-3 \vec{b}+2 \vec{b} \Rightarrow 2 \vec{c}=3 \vec{a}-\vec{b} \\ \therefore \quad \vec{c} =\frac{3 \vec{a}-\vec{b}}{2} \end{matrix} $

Hence, the required vector is $\vec{c}=\frac{3 \vec{a}-\vec{b}}{2}$.

Solution

Let the given points are $A(k,-10,3), B(1,-1,3)$ and $C(3,5,3)$

$ \begin{aligned} \overrightarrow{{}AB} & =(1-k) \hat{i}+(-1+10) \hat{j}+(3-3) \hat{k} \\ \overrightarrow{{}AB} & =(1-k) \hat{i}+9 \hat{j}+0 \hat{k} \\ \therefore \mid \overrightarrow{{}AB} & =\sqrt{(1-k)^{2}+(9)^{2}}=\sqrt{(1-k)^{2}+81} \\ \overrightarrow{{}BC} & =(3-1) \hat{i}+(5+1) \hat{j}+(3-3) \hat{k}=2 \hat{i}+6 \hat{j}+0 \hat{k} \\ \therefore \mid \overrightarrow{{}BC} & =\sqrt{(2)^{2}+(6)^{2}}=\sqrt{4+36}=\sqrt{40}=2 \sqrt{10} \\ \overrightarrow{{}AC} & =(3-k) \hat{i}+(5+10) \hat{j}+(3-3) \hat{k}=(3-k) \hat{i}+15 \hat{j}+0 \hat{k} \\ \therefore \quad|\overrightarrow{{}AC}| & =\sqrt{(3-k)^{2}+(15)^{2}}=\sqrt{(3-k)^{2}+225} \end{aligned} $

If $A, B$ and $C$ are collinear, then

$ \begin{aligned} |\overrightarrow{{}AB}|+|\overrightarrow{{}BC}| & =|\overrightarrow{{}AC}| \\ \sqrt{(1-k)^{2}+81}+\sqrt{40} & =\sqrt{(3-k)^{2}+225} \end{aligned} $

Squaring both sides, we have

${[\sqrt{(1-k)^{2}+81}+\sqrt{40}]^{2}=[\sqrt{(3-k)^{2}+225}]^{2}}$

$\Rightarrow (1-k)^{2}+81+40+2 \sqrt{40} \sqrt{(1-k)^{2}+81}=(3-k)^{2}+225$

$\Rightarrow 1+k^{2}-2 k+121+2 \sqrt{40} \sqrt{1+k^{2}-2 k+81} =9+k^{2}-6 k+225$

$\Rightarrow 122-2 k+2 \sqrt{40} \sqrt{k^{2}-2 k+82}=234-6 k$

Dividing by 2 , we get

$ \begin{matrix} \Rightarrow & 61-k+\sqrt{40} \sqrt{k^{2}-2 k+82}=117-3 k \\ \Rightarrow & \sqrt{40} \sqrt{k^{2}-2 k+82}=117-61-3 k+k \\ \Rightarrow & \sqrt{40} \sqrt{k^{2}-2 k+82}=56-2 k \Rightarrow 2 \sqrt{10} \sqrt{k^{2}-2 k+82}=56-2 k \\ \Rightarrow & \sqrt{10} \sqrt{k^{2}-2 k+82}=28-k \quad \text{ (Dividing by } 2 \text{ ) } \end{matrix} $

Squaring both sides, we get

$ \begin{aligned} & \Rightarrow \quad 10(k^{2}-2 k+82)=784+k^{2}-56 k \\ & \Rightarrow \quad 10 k^{2}-20 k+820=784+k^{2}-56 k \\ & \Rightarrow \quad 10 k^{2}-k^{2}-20 k+56 k+820-784=0 \\ & \Rightarrow 9 k^{2}+36 k+36=0 \Rightarrow k^{2}+4 k+4=0 \Rightarrow(k+2)^{2}=0 \end{aligned} $

$ \Rightarrow \quad k+2=0 \quad \Rightarrow k=-2 $

Hence, the required value is $k=-2$

Solution

Since, the vector $\vec{r}$ makes equal angles with the axes, their direction cosines should be same

$\therefore \quad l=m=n$

We know that $\quad l^{2}+m^{2}+n^{2}=1 \quad \Rightarrow \quad l^{2}+l^{2}+l^{2}=1$

$ \begin{matrix} \Rightarrow \quad 3 l^{2}=1 \Rightarrow l^2=\frac{1}{3} \Rightarrow l=\pm \frac{1}{\sqrt {3}} \\ \quad \hat{r}= \pm \frac{1}{\sqrt{3}} \hat{ i }\pm \frac{1}{\sqrt{3}} \hat{j} \pm \frac{1}{\sqrt{3}} \hat{k} \Rightarrow \hat{r}= \pm \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) \\ \end{matrix} $

We know that $\vec{r}=(\hat{r})|\vec{r}|$

$ = \pm \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) 2 \sqrt{3}= \pm 2(\hat{i}+\hat{j}+\hat{k}) $

Hence, the required value of $\vec{r}$ is $\pm 2(\hat{i}+\hat{j}+\hat{k})$.

Solution

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a}=2 k, \vec{b}=3 k$ and $\vec{c}=-6 k$

If $l, m$ and $n$ are the direction cosines of vector $\vec{r}$, then

$ \begin{aligned} l & =\frac{\vec{a}}{|\vec{r}|}=\frac{2 k}{14}=\frac{k}{7} \\ m & =\frac{\vec{b}}{|\vec{r}|}=\frac{3 k}{14} \text{ and } n=\frac{\vec{c}}{|\vec{r}|}=\frac{-6 k}{14}=\frac{-3 k}{7} \end{aligned} $

We know that $l^{2}+m^{2}+n^{2}=1$

$ \begin{matrix} \therefore & \frac{k^{2}}{49}+\frac{9 k^{2}}{196}+\frac{9 k^{2}}{49}=1 \\ \Rightarrow & \frac{4 k^{2}+9 k^{2}+36 k^{2}}{196}=1 \Rightarrow 49 k^{2}=196 \Rightarrow k^{2}=4 \\ \therefore & k= \pm 2 \text{ and } l=\frac{k}{7}=\frac{2}{7} \\ & m=\frac{3 k}{14}=\frac{3 \times 2}{14}=\frac{3}{7} \text{ and } n=\frac{-3 k}{7} \frac{-3 \times 2}{7}=\frac{-6}{7} \\ & \hat{r}= \pm(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}) \end{matrix} $

$ \begin{aligned} \hat{r} & =\hat{r}|\vec{r}| \\ \Rightarrow \quad \vec{r} & = \pm(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}) \cdot 14= \pm(4 \hat{i}+6 \hat{j}-12 \hat{k}) \end{aligned} $

Hence, the required direction cosines are $\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}$ and the components of $\vec{r}$ are $4 \hat{i}, 6 \hat{j}$ and $-12 \hat{k}$.

Solution

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

We know that unit vector perpendicular to $\vec{a}$ and $\vec{b}=\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}$

$ \begin{aligned} \vec{a} \times \vec{b} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{vmatrix} \\ & =\hat{i}(-3+2)-\hat{j}(6-8)+\hat{k}(-2+4)=-\hat{i}+2 \hat{j}+2 \hat{k} \end{aligned} $

$\therefore \quad|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(2)^{2}+(2)^{2}}=\sqrt{1+4+4}=\sqrt{9}=3$

so, $\quad \frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}=\frac{-\hat{i}+2 \hat{j}+2 \hat{k}}{3}=\frac{1}{3}(-\hat{i}+2 \hat{j}+2 \hat{k})$

Now the vector of magnitude $6=\frac{1}{3}(-\hat{i}+2 \hat{j}+2 \hat{k}) \cdot 6$

$ =2(-\hat{i}+2 \hat{j}+2 \hat{k})=-2 \hat{i}+4 \hat{j}+4 \hat{k} $

Hence, the required vector is $-2 \hat{i}+4 \hat{j}+4 \hat{k}$.

Solution

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

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

$ \begin{aligned} \therefore \quad \cos \theta & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(3 \hat{i}+4 \hat{j}-\hat{k})}{\sqrt{4+1+1} \cdot \sqrt{9+16+1}} \\ & =\frac{6-4-1}{\sqrt{6} \cdot \sqrt{26}} \Rightarrow \frac{1}{2 \sqrt{3} \cdot \sqrt{13}}=\frac{1}{2 \sqrt{39}} \\ \therefore \quad \theta & =\cos ^{-1} \frac{1}{2 \sqrt{39}} \Rightarrow \theta=\cos ^{-1}(\frac{1}{156}) \end{aligned} $

Hence, the required value of $\theta$ is $\cos ^{-1}(\frac{1}{156})$.

Solution

Given that $\vec{a}+\vec{b}+\vec{c}=0$

$\text{So,} \qquad \vec{a} \times(\vec{a}+\vec{b}+\vec{c}) =\vec{a} \times 0$

$\Rightarrow \qquad \vec{a} \times \vec{a}+\vec{a} \times \vec{b}+\vec{a} \times \vec{c} =0 $

$\Rightarrow \qquad \vec{o}+\vec{a} \times \vec{b}+\vec{a} \times \vec{c}=0 \qquad (\vec{a} \times \vec{a}=0)$

$\Rightarrow \qquad \vec{a} \times \vec{b}-\vec{c} \times \vec{a}=0 \qquad (\vec{a} \times \vec{c}=-\vec{c} \times \vec{a})$

$\Rightarrow \qquad \vec{a} \times \vec{b}=\vec{c} \times \vec{a} \qquad \ldots(i)$

Now $\qquad \vec{a}+\vec{b}+\vec{c}=0$

$\Rightarrow \qquad \vec{b} \times(\vec{a}+\vec{b}+\vec{c})=\vec{b} \times 0$

$\Rightarrow \qquad \vec{b} \times \vec{a}+\vec{b} \times \vec{b}+\vec{b} \times \vec{c}=0$

$\Rightarrow \qquad \vec{b} \times \vec{a}+\vec{o}+\vec{b} \times \vec{c}=0 \qquad (\because \quad \vec{b} \times \vec{b}=0)$

$\Rightarrow \qquad -(\vec{a} \times b)+\vec{b} \times \vec{c} =0$

$\Rightarrow \qquad \vec{b} \times \vec{c} =\vec{a} \times \vec{b}$

From eq. (i) and (ii) we get $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$. Hence proved.

Geometrical Interpretation

According to figure, we have

Area of parallelogram ABCD is

$\Rightarrow|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$

Since, the parallelograms on the same base and between the same parallel lines are equal in area

$\therefore|\vec{a} \times \vec{b}|=|\vec{b} \times \vec{c}|=|\vec{c} \times \vec{a}|$

$\Rightarrow \vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$.

Solution

Given that $\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}$

We know that $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$

$ \begin{aligned} \therefore \quad \vec{a} \times \vec{b} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4 \end{vmatrix} \\ & =\hat{i}(4+4)-\hat{j}(12-4)+\hat{k}(-6-2) \\ & =8 \hat{i}-8 \hat{j}-8 \hat{k} \\ |\vec{a} \times \vec{b}| & =\sqrt{(8)^{2}+(-8)^{2}+(-8)^{2}} \end{aligned} $

$ \begin{aligned} & =\sqrt{64+64+64}=\sqrt{192}=\sqrt{64 \times 3}=8 \sqrt{3} \\ |\vec{a}| & =\sqrt{(3)^{2}+(1)^{2}+(2)^{2}}=\sqrt{9+1+4}=\sqrt{14} \\ |\vec{b}| & =\sqrt{(2)^{2}+(-2)^{2}+(4)^{2}}=\sqrt{4+4+16} \\ & =\sqrt{24}=2 \sqrt{6} \\ \therefore \quad \sin \theta & =\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{8 \sqrt{3}}{\sqrt{14} \cdot 2 \sqrt{6}} \\ \Rightarrow \quad & \frac{4 \sqrt{3}}{\sqrt{84}}=\frac{4 \sqrt{3}}{2 \sqrt{21}}=\frac{2}{\sqrt{7}} \\ \text{ Hence, } \sin \theta & =\frac{2}{\sqrt{7}} . \end{aligned} $

Solution

Here, Position vector of $A=\hat{i}+\hat{j}-\hat{k}$

Position vector of $B=2 \hat{i}-\hat{j}+3 \hat{k}$

Position vector of $C=2 \hat{i}-3 \hat{k}$

Position vector of $D=3 \hat{i}-2 \hat{j}+\hat{k}$

$\overrightarrow{{}AB}=P . V$ of $B-P . V$ of $A$ $=(2 \hat{i}-\hat{j}+3 \hat{k})-(\hat{i}+\hat{j}-\hat{k})=\hat{i}-2 \hat{j}+4 \hat{k}$

$\overrightarrow{{}CD}=$ P.V. of D - P.V. of C $=(3 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}-3 \hat{k})=\hat{i}-2 \hat{j}+4 \hat{k}$

Projection of $\overrightarrow{{}AB}$ on $\overrightarrow{{}CD}=\frac{\overrightarrow{{}AB} \cdot \overrightarrow{{}CD}}{|\overrightarrow{{}CD}|}$

$ \begin{aligned} & =\frac{(\hat{i}-2 \hat{j}+4 \hat{k}) \cdot(\hat{i}-2 \hat{j}+4 \hat{k})}{\sqrt{(1)^{2}+(-2)^{2}+(4)^{2}}} \\ & =\frac{1+4+16}{\sqrt{1+4+16}}=\frac{21}{\sqrt{21}}=\sqrt{21} \end{aligned} $

Hence, the required projection $=\sqrt{21}$.

Solution

Given that $A(1,2,3), B(2,-1,4)$ and $C(4,5,-1)$

$ \begin{aligned} & \qquad \begin{aligned} & \overrightarrow{{}AB}=(2-1) \hat{i}+(-1-2) \hat{j}+(4-3) \hat{k} \\ & \begin{aligned} \overrightarrow{{}AB}=\hat{i}-3 \hat{j}+\hat{k} \end{aligned} \\ & \begin{aligned} \overrightarrow{{}AC}=(4-1) \hat{i}+(5-2) \hat{j}+(-1-3) \hat{k}=3 \hat{i}+3 \hat{j}-4 \hat{k} \end{aligned} \\ & \text{ Area of } \triangle ABC=\frac{1}{2}|\overrightarrow{{}AB} \times \overrightarrow{{}AC}|=\frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix} \\ &=\frac{1}{2}[\hat{i}(12-3)-\hat{j}(-4-3)+\hat{k}(3+9)] \\ &=\frac{1}{2}|9 \hat{i}+7 \hat{j}+12 \hat{k}|=\frac{1}{2} \sqrt{(9)^{2}+(7)^{2}+(12)^{2}} \\ &=\frac{1}{2} \sqrt{81+49+144}=\frac{1}{2} \sqrt{274} \end{aligned} \end{aligned} $

Hence, the required area is $\frac{\sqrt{274}}{2}$.

Solution

Let $A B C D$ and $A B F E$ be two parallelograms on the same base $AB$ and between same parallel lines $AB$ and $DF$.

Let

$ \overrightarrow{{}AB}=\vec{a} \text{ and } \overrightarrow{{}AD}=\vec{b} $

$\therefore \quad$ Area of parallelogram $ABCD=|\vec{a} \times \vec{b}|$

Now Area of parallelogram $ABFE=|\overrightarrow{{}AB} \times \overrightarrow{{}AE}|$

$ \begin{aligned} & =|\vec{a} \times(\overrightarrow{{}AD}+\overrightarrow{{}DE})|=|\vec{a} \times(\vec{b} \times K \vec{a})| \\ & =|(\vec{a} \times \vec{b})+K(\vec{a} \times \vec{a})|=|\vec{a} \times \vec{b}|+0 \quad[\because \vec{a} \times \vec{a}=0] \\ & =|\vec{a} \times \vec{b}| \end{aligned} $

Hence proved.

Solution

Here, in the given figure, the components of $c$ are $c \cos A$ and $c \sin A$.

$\therefore \quad \overrightarrow{{}CD}=b-c \cos A$

In $\triangle BDC$,

$ a^{2}=CD^{2}+BD^{2} $

$\Rightarrow \quad a^{2}=(b-c \cos A)^{2}+(c \sin A)^{2}$

$\Rightarrow \quad a^{2}=b^{2}+c^{2} \cos ^{2} A-2 b c \cos A+c^{2} \sin ^{2} A$

$\Rightarrow \quad a^{2}=b^{2}+c^{2}(\cos ^{2} A+\sin ^{2} A)-2 b c \cos A$

$\Rightarrow \quad a^{2}=b^{2}+c^{2}-2 b c \cos A \Rightarrow 2 b c \cos A=b^{2}+c^{2}-a^{2}$

$\therefore \quad \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$

Hence Proved.

Solution

Since, $\vec{a}, \vec{b}$ and $\vec{c}$ are the vertices of $\triangle ABC$

$\therefore \quad \overrightarrow{{}AB}=\vec{b}-\vec{a}, \overrightarrow{{}BC}=\bar{{}c}-\vec{b}$

and $\overrightarrow{{}AC}=\vec{c}-\vec{a}$

$\therefore \quad$ Area of $\triangle ABC=\frac{1}{2}|\overrightarrow{{}AB} \times \overrightarrow{{}AC}|$

$=\frac{1}{2}|(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})|$

$=\frac{1}{2}|\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{a} \times \vec{c}+\vec{a} \times \vec{a}|$

$=\frac{1}{2}|\vec{b} \times \vec{c}+\vec{a} \times \vec{b}+\vec{c} \times \vec{a}|$

$ \begin{bmatrix} \because & \vec{a} \times \vec{b}=-\vec{b} \times \vec{a} \\ & \vec{c} \times \vec{a}=-\vec{a} \times \vec{c} \\ & \vec{a} \times \vec{a}=\overrightarrow{{}0} \end{bmatrix} $

For three vectors are collinear, area of $\triangle ABC=0$

$ \begin{aligned} \therefore \quad \frac{1}{2}|\vec{b} \times \vec{c}+\vec{a} \times \vec{b}+\vec{c} \times \vec{a}| & =0 \\ |\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}| & =0 \end{aligned} $

which is the condition of collinearity of $\vec{a}, \vec{b}$ and $\vec{c}$.

Let $\hat{n}$ be the unit vector normal to the plane of the $\triangle ABC$

$ \begin{aligned} & \therefore \quad \hat{n}=\frac{\overrightarrow{{}AB} \times \overrightarrow{{}AC}}{|\overrightarrow{{}AB} \times \overrightarrow{{}AC}|} \\ & \Rightarrow \quad=\frac{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}}{|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|} \end{aligned} $

Solution

Let $ABCD$ be a parallelogram such that,

$ \overrightarrow{{}AB}=\vec{p}, \overrightarrow{{}AD}=\vec{q}=\overrightarrow{{}BC} $

$\therefore$ by law of triangle, we get

$$ \begin{align*} \overrightarrow{{}AC} & =\vec{a}=\vec{p}+\vec{q} \tag{i}\\ \text{ and } \overrightarrow{{}BD} & =\vec{b}=-\vec{p}+\vec{q} \tag{ii} \end{align*} $$

Adding eq. (i) and (ii) we get,

$ \vec{a}+\vec{b}=2 \vec{q} \quad \Rightarrow \quad \vec{q}=(\frac{\vec{a}+\vec{b}}{2}) $

Subtracting eq. (ii) from eq. (i) we get

$ \begin{aligned} \vec{a}-\vec{b} & =2 \vec{p} \Rightarrow \vec{p}=(\frac{\vec{a}-\vec{b}}{2}) \\ \therefore \vec{p} \times \vec{q} & =\frac{1}{4}(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\frac{1}{4}(\vec{a} \times \vec{a}-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}-\vec{b} \times \vec{b}) \\ & =\frac{1}{4}(-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}) \\ & =\frac{1}{4}(\vec{a} \times \vec{b}+\vec{a} \times \vec{b})=\frac{1}{4} \cdot 2(\vec{a} \times \vec{b})=\frac{|\vec{a} \times \vec{b}|}{2} \end{aligned} $

So, the area of the parallelogram ABCD $=|\vec{p} \times \vec{q}|=\frac{1}{2}|\vec{a} \times \vec{b}|$ Now area of parallelogram whose diagonals are $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3 \hat{j}-\hat{k}=\frac{1}{2}|(2 \hat{i}-\hat{j}+\hat{k}) \times(\hat{i}+3 \hat{j}-\hat{k})|$

$ =- \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{vmatrix} $

$ \begin{aligned} & =\frac{1}{2}|\hat{i}(1-3)-\hat{j}(-2-1)+\hat{k}(6+1)|=\frac{1}{2}|-2 \hat{i}+3 \hat{j}+7 \hat{k}| \\ & =\frac{1}{2} \sqrt{(-2)^{2}+(3)^{2}+(7)^{2}}=\frac{1}{2} \sqrt{4+9+49} \\ & =\frac{1}{2} \sqrt{62} \text{ sq. units } \end{aligned} $

Hence, the required area is $\frac{1}{2} \sqrt{62}$ sq. units.

Solution

Let $\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$

Also given that $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}$

Since,

$ \vec{a} \times \vec{c}=\vec{b} $

$\therefore \quad \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ c_1 & c_2 & c_3\end{vmatrix} =\hat{j}-\hat{k}$

$ =\hat{i}(c_3-c_2)-\hat{j}(c_3-c_1)+\hat{k}(c_2-c_1)=\hat{j}-\hat{k} $

On comparing the like terms, we get

$ \begin{aligned} & c_3-c_2=0 \quad …(i)\\ & c_1-c_3=1 \quad …(ii)\\ & \text{ and } \quad c_2-c_1=-1 \quad …(iii) \\ & \text{ for } \vec{a} \cdot \vec{c}=3 \\ & (\hat{i}+\hat{j}+\hat{k}) \cdot(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k})=3 \\ & \therefore \quad c_1+c_2+c_3=3 \quad …(iv) \end{aligned} $

Adding eq. (ii) and eq. (iii) we get,

From (iv) and $(v)$ we get

$c_2-c_3=0 \quad …(v)$

From (iii) and (vi) we get

$c_1+2 c_2=3 \quad …(vi)$

$ \begin{gathered} c_1+2 c_2=3 \\ -c_1+c-2=-1 \\ \hline 3 c_2=2 \\ \therefore c_2=\frac{2}{3} \\ c_3-c_2=0 \Rightarrow c_3-\frac{2}{3}=0 \end{gathered} $

$ \therefore \quad c_3=\frac{2}{3} $

Now $\quad c_2-c_1=-1 \Rightarrow \frac{2}{3}-c_1=-1$

$\Rightarrow \quad c_1=1+\frac{2}{3}=\frac{5}{3}$

$\therefore \quad \vec{c}=\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}$

Hence, $\quad \vec{c}=\frac{1}{3}(5 \hat{i}+2 \hat{j}+2 \hat{k})$.

Solution

Let $\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}$

Unit vector in the direction of $\vec{a}=\frac{\vec{a}}{|\vec{a}|}$

$ =\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{\sqrt{(1)^{2}+(-2)^{2}+(2)^{2}}}=\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{\sqrt{1+4+4}}=\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3} $

$\therefore \quad$ Vector of magnitude $9=\frac{9(\hat{i}-2 \hat{j}+2 \hat{k})}{3}=3(\hat{i}-2 \hat{j}+2 \hat{k})$

Hence, the correct option is (c).

Solution

The given vectors are $2 \vec{a}-3 \vec{b}$ and $\vec{a}+\vec{b}$ and the ratio is $3: 1$. $\therefore \quad$ The position vector of the required point $c$ which divides the join of the given vectors $\vec{a}$ and $\vec{b}$ is

$ \vec{c}=\frac{m_1 x_2+m_2 x_1}{m_1+m_2} $

$ \begin{aligned} & =\frac{1 \cdot(2 \vec{a}-3 \vec{b})+3(\vec{a}+\vec{b})}{3+1}=\frac{2 \vec{a}-3 \vec{b}+3 \vec{a}+3 \vec{b}}{4} \\ & =\frac{5 \vec{a}}{4}=\frac{5}{4} \vec{a} \end{aligned} $

Hence, the correct option is (d).

Solution

Let A and B be two points whose coordinates are given as $(2,5,0)$ and $(-3,7,4)$

$ \begin{matrix} \therefore & \overrightarrow{{}AB}=(-3-2) \hat{i}+(7-5) \hat{j}+(4-0) \hat{k} \\ \Rightarrow & \overrightarrow{{}AB}=-5 \hat{i}+2 \hat{j}+4 \hat{k} \end{matrix} $

Hence, the correct option is (c).

Solution

Here, given that $|\vec{a}|=\sqrt{3},|\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2 \sqrt{3}$

$\therefore$ From scalar product, we know that

$ \begin{aligned} & \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\ & \Rightarrow \quad 2 \sqrt{3}=\sqrt{3} \cdot 4 \cdot \cos \theta \\ & \Rightarrow \quad \cos \theta=\frac{2 \sqrt{3}}{\sqrt{3} \cdot 4}=\frac{1}{2} \\ & \therefore \quad \theta=\frac{\pi}{3} \end{aligned} $

Hence, the correct option is (b).

Solution

Given that $\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$

Since $\vec{a}$ and $\vec{b}$ are orthogonal

$ \begin{matrix} \therefore & \vec{a} \cdot \vec{b}=0 \\ \Rightarrow & (2 \hat{i}+\lambda \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=0 \end{matrix} $

$ \begin{aligned} \Rightarrow & 2+2 \lambda+3 & =0 \\ \Rightarrow & 5+2 \lambda & =0 \Rightarrow \lambda=\frac{-5}{2} \end{aligned} $

Hence, the correct option is (d).

Solution

Let

$ \begin{aligned} & \vec{a}=3 \hat{i}-6 \hat{j}+\hat{k} \\ & \vec{b}=2 \hat{i}-4 \hat{j}+\lambda \hat{k} \end{aligned} $

Since the given vectors are parallel,

$\therefore$ Angle between them is $0^{\circ}$

$ \begin{aligned} & \text{ so } \quad \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos 0 \\ & \Rightarrow(3 \hat{i}-6 \hat{j}+\hat{k}) \cdot(2 \hat{i}-4 \hat{j}+\lambda \hat{k})=|3 \hat{i}-6 \hat{j}+\hat{k}||2 \hat{i}-4 \hat{j}+\lambda \hat{k}| \\ & 6+24+\lambda=\sqrt{9+36+1} \cdot \sqrt{4+16+\lambda^{2}} \\ & 30+\lambda=\sqrt{46} \cdot \sqrt{20+\lambda^{2}} \end{aligned} $

Squaring both sides, we get

$900+\lambda^{2}+60 \lambda =46(20+\lambda^{2})$

$\Rightarrow 900+\lambda^{2}+60 \lambda =920+46 \lambda^{2}$

$\Rightarrow \lambda^{2}-46 \lambda^{2}+60 \lambda+900-920 =0$

$\Rightarrow -45 \lambda^{2}+60 \lambda-20 =0$

$\Rightarrow 9 \lambda^{2}-12 \lambda+4 =0$

$\Rightarrow (3 \lambda-2)^{2} =0$

$\Rightarrow 3 \lambda-2 =0$

$\Rightarrow 3 \lambda =2$

$\therefore \lambda =2 / 3$

Alternate method

Let $\quad \vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \quad \text{ and } \quad \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$

If $\quad \vec{a} | \vec{b}$

$ \begin{matrix} \therefore \qquad \frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3} \\ \Rightarrow \qquad \frac{3}{2}=\frac{-6}{-4}=\frac{1}{\lambda} \Rightarrow \frac{1}{\lambda}=\frac{3}{2} \Rightarrow \lambda=\frac{2}{3} \end{matrix} $

Hence, the correct option is (a).

Solution

Let $O$ be the origin

$ \begin{aligned} & \therefore \quad \overrightarrow{{}OA}=2 \hat{i}-3 \hat{j}+2 \hat{k} \\ & \overrightarrow{{}OB}=2 \hat{i}+3 \hat{j}+\hat{k} \end{aligned} $

$\therefore \quad$ Area of $\triangle OAB=\frac{1}{2}|\overrightarrow{{}OA} \times \overrightarrow{{}OB}|=\frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 2 \\ 2 & 3 & 1\end{vmatrix} $

$ \begin{aligned} & =\frac{1}{2}|\hat{i}(-3-6)-\hat{j}(2-4)+\hat{k}(6+6)| \\ & =\frac{1}{2}|-9 \hat{i}+2 \hat{j}+12 \hat{k}| \\ & =\frac{1}{2} \sqrt{(-9)^{2}+(2)^{2}+(12)^{2}} \\ & =\frac{1}{2} \sqrt{81+4+144}=\frac{1}{2} \sqrt{229} \end{aligned} $

Hence the correct option is $(d)$.

Solution

Let

$ \vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} $

$\therefore \quad \vec{a} ^{2}=a_1^{2}+a_2^{2}+a_3^{2}$

Now, $\quad \vec{a} \times \hat{i}=(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}) \times \hat{i}$

$ \begin{aligned} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 1 & 0 & 0 \end{vmatrix} \\ & =\hat{i}(0-0)-\hat{j}(0-a_3)+\hat{k}(0-a_2)=a_3 \hat{j}-a_2 \hat{k} \\ \therefore \quad(\vec{a} \times \hat{i})^{2} & =(a_3 \hat{j}-a_2 \hat{k}) \cdot(a_3 \hat{j}-a_2 \hat{k})=a_3^{2}+a_2^{2} \end{aligned} $

Similarly

$ (\vec{a} \times \hat{j})^{2}=a_1^{2}+a_3^{2} $

and

$ (\vec{a} \times \hat{k})^{2}=a_1^{2}+a_2^{2} $

$ \begin{aligned} \therefore(\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2} & =a_3^{2}+a_2^{2}+a_1^{2}+a_3^{2}+a_1^{2}+a_2^{2} \\ & =2(a_1^{2}+a_2^{2}+a_3^{2})=2 \vec{a} ^{2} \end{aligned} $

Hence, the correct option is (d).

Solution

Given that $|\vec{a}|=10,|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=12$

$ \begin{aligned} & \therefore \quad \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\ & \Rightarrow \quad 12=10 \cdot 2 \cdot \cos \theta \\ & \Rightarrow \quad \cos \theta=\frac{12}{20}=\frac{3}{5} \\ & \therefore \quad \sin \theta=\sqrt{1-\cos ^{2} \theta} \\ & \Rightarrow \quad \sin \theta=\sqrt{1-(\frac{3}{5})^{2}} \Rightarrow \sin \theta=\sqrt{1-\frac{9}{25}} \\ & \Rightarrow \quad \sin \theta=\sqrt{\frac{16}{25}} \quad \Rightarrow \sin \theta=\frac{4}{5} \end{aligned} $

Now $\quad|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$

$ =10 \cdot 2 \cdot \frac{4}{5}=16 $

Hence, the correct option is $(d)$.

Solution

Let

$ \begin{aligned} \vec{a} & =\lambda \hat{i}+\hat{j}+2 \hat{k} \\ \vec{b} & =\hat{i}+\lambda \hat{j}-\hat{k} \\ \vec{c} & =2 \hat{i}-\hat{j}+\lambda \hat{k} \end{aligned} $

If $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar, then

$\vec{a} \cdot(\vec{b} \times \vec{c})=0$

$\therefore \qquad \begin{vmatrix} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{vmatrix} =0$

$\Rightarrow \lambda(\lambda^{2}-1)-1(\lambda+2)+2(-1-2 \lambda) =0$

$\Rightarrow \lambda^{3}-\lambda-\lambda-2-2-4 \lambda =0$

$\Rightarrow \lambda^{3}-6 \lambda-4 =0$

$\Rightarrow (\lambda-2)(\lambda^{2}-2 \lambda-2) =0$

$\Rightarrow \lambda =-2 \text{ or } \lambda^{2}-2 \lambda-2 =0$

$ \begin{matrix} \Rightarrow & \lambda=\frac{2 \pm \sqrt{4+8}}{2} \\ \Rightarrow & \lambda=\frac{2 \pm 2 \sqrt{3}}{2} \\ \therefore & \lambda=-2 \text{ or } \lambda=1 \pm \sqrt{3} \end{matrix} $

Hence, the correct option is $(a)$.

Solution

Given that $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$

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

$ \begin{aligned} & \therefore \quad(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=\overrightarrow{{}0} \cdot \overrightarrow{{}0}=0 \\ & |\vec{a}|^{2}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{a}+|\vec{b}|^{2}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+|\vec{c}|^{2}=0 \\ & \Rightarrow \quad|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a}=0 \\ & \Rightarrow \quad 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot a)=0 \\ & \Rightarrow \quad 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-3 \\ & \therefore \quad \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2} \end{aligned} $

Hence, the correct option is (c).

Solution

The projection vector of $\vec{a}$ on $\vec{b}=(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}) \cdot \vec{b}$ Hence, the correct option is (a).

Solution

Given that $|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=5$,

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

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

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

$\Rightarrow \quad|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a}=0$

$\Rightarrow (2)^{2}+(3)^{2}+(5)^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$

$\Rightarrow 4+9+25+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$

$\Rightarrow 38+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$

$\Rightarrow 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-38$

$\therefore \qquad \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-19$

Hence, the correct option is (c).

Solution

Given that $|\vec{a}|=4,-3 \leq \lambda \leq 2$

Now $|\lambda \vec{a}|=\lambda|\vec{a}|=\lambda \cdot 4=4 \lambda$

Here $\quad-3 \leq \lambda \leq 2$

$\Rightarrow-3.4 \leq 4 \lambda \leq 2.4 \Rightarrow-12 \leq 4 \lambda \leq 8$

$\therefore \quad 4 \lambda=[-12,8]$

Hence, the correct option is (b).

Solution

The number of vectors of unit length perpendicular to vectors $\vec{a}$ and $\vec{b}$ is $\vec{c}$ (let)

$\therefore \quad \vec{c}= \pm(\vec{a} \times \vec{b})$

So, there will be two vectors of unit length perpendicular to vectors $\vec{a}$ and $\vec{b}$.

Hence, the correct option is $(b)$.

Solution

If vector $\vec{a}+\vec{b}$ bisects the angle between non-collinear vectors $\vec{a}$ and $\vec{b}$ then the angle between $\vec{a}+\vec{b}$ and $\vec{a}$ is equal to the angle between $\vec{a}+\vec{b}$ and $\vec{b}$.

So,

$$ \begin{align*} \cos \theta & =\frac{\vec{a} \cdot(\vec{a}+\vec{b})}{|\vec{a}||\vec{a}+\vec{b}|}=\frac{\vec{a} \cdot(\vec{a}+\vec{b})}{|\vec{a}| \sqrt{a^{2}+b^{2}}} \tag{i}\\ \text{Also,} \qquad \cos \theta & =\frac{\vec{b} \cdot (\vec{a}+\vec{b})}{|\vec{b}| \cdot|\vec{a}+\vec{b}|} \quad [\therefore \theta \text{ is same }] \\ & =\frac{\vec{b} \cdot(\vec{a}+\vec{b})}{|\vec{b}| \sqrt{a^{2}+b^{2}}} \tag{ii} \end{align*} $$

From eq. (i) and eq. (ii) we get,

$\frac{\vec{a} \cdot(\vec{a}+\vec{b})}{|\vec{a}| \sqrt{a^{2}+b^{2}}} =\frac{\vec{b} \cdot(\vec{a}+\vec{b})}{|\vec{b}| \sqrt{a^{2}+b^{2}}}$

$\Rightarrow \qquad \frac{\vec{a}}{|\vec{a}|} =\frac{\vec{b}}{|\vec{b}|}$

$\Rightarrow \qquad \hat{a} =\hat{b} \Rightarrow \vec{a}=\vec{b}$

Hence, the required filler is $\vec{a}=\vec{b}$.

Solution

If $\vec{r}$ is a non-zero vector, then $\vec{a}, \vec{b}$ and $\vec{c}$ can be in the same plane.

Since angles between $\vec{a}, \quad$ and $\vec{c}$ are zero i.e. $\theta=0$

$\therefore \quad \vec{a} \cdot(\vec{b} \times \vec{c})=0$

Hence the required value is 0 .

Solution

Given that $\quad \vec{a}=3 \hat{i}-2 \hat{j}+2 \hat{k}$

and $\quad \vec{b}=-\hat{i}-2 \hat{k}$

$\therefore \quad \vec{a}+\vec{b}=2 \hat{i}-2 \hat{j}$ and $\vec{a}-\vec{b}=4 \hat{i}-2 \hat{j}+4 \hat{k}$

Let $\theta$ be the angle between the two diagonal vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ then

$ \begin{aligned} \cos \theta & =\frac{(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})}{|\vec{a}+\vec{b}||\vec{a}-\vec{b}|}=\frac{(2 \hat{i}-2 \hat{j}) \cdot(4 \hat{i}-2 \hat{j}+4 \hat{k})}{\sqrt{(2)^{2}+(-2)^{2}} \cdot \sqrt{(4)^{2}+(-2)^{2}+(4)^{2}}} \\ & =\frac{8+4}{2 \sqrt{2} \cdot 6}=\frac{12}{2 \sqrt{2} \cdot 6}=\frac{1}{\sqrt{2}} \end{aligned} $

$ \therefore \quad \theta=\frac{\pi}{4} $

Hence the value of required filler is $\frac{\pi}{4}$.

Solution

Given that $|k \vec{a}|<|\vec{a}|$ and $k \vec{a}+\frac{1}{2} \vec{a}$ is parallel to $\vec{a}$

$ \therefore|k \vec{a}|<|\vec{a}| \Rightarrow|k||\vec{a}|<|\vec{a}| \Rightarrow|k|<1 \Rightarrow-1<k<1 $

Now since $k \vec{a}+\frac{1}{2} \vec{a}$ is parallel to $\vec{a}$

Here we see that at $k=-\frac{1}{2}, k \vec{a}+\frac{1}{2} \vec{a}$ become null vector and then it will not be parallel to $\vec{a}$.

$\therefore k \vec{a}+\frac{1}{2} \vec{a}$ is parallel to $\vec{a}$ when $k \in(-1,1)$ and $k \neq \frac{1}{2}$.

Hence, the required value of $k \in(-1,1)$ and $k \neq \frac{1}{2}$.

Solution

$ \begin{aligned} |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2} & =(|\vec{a}||\vec{b}| \sin \theta)^{2}+(|\vec{a}||\vec{b}| \cos \theta)^{2} \\ & =|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \\ & =|\vec{a}|^{2}|\vec{b}|^{2} \cdot(\sin ^{2} \theta+\cos ^{2} \theta) \\ & =|\vec{a}|^{2}|\vec{b}|^{2} \cdot 1=|\vec{a}|^{2}|\vec{b}|^{2} \end{aligned} $

Hence, the value of the filler is $|\vec{a}|^{2}|\vec{b}|^{2}$.

Solution

$ \begin{aligned} & r l & |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2} & =144 \\ \Rightarrow & & (|\vec{a}||\vec{b}| \sin \theta)^{2}+(|\vec{a}||\vec{b}| \cos \theta)^{2} & =144 \\ & \Rightarrow & |\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta & =144 \\ \Rightarrow & & |\vec{a}|^{2}|\vec{b}|^{2}(\sin ^{2} \theta+\cos ^{2} \theta) & =144 \\ & \Rightarrow & |\vec{a}|^{2}|\vec{b}|^{2} & =144 \\ \Rightarrow & & |\vec{a}||\vec{b}| & =12 \\ & \therefore & 4 \cdot|\vec{b}| & =12 \\ & & |\vec{b}| & =3 \end{aligned} $

Hence, the value of the filler is 3 .

Solution

Let

$ \begin{aligned} \vec{a} & =a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \\ \cdot \hat{i} & =(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}. \\ & =a_1 \end{aligned} $

$ \therefore \quad \vec{a} \cdot \hat{i}=(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}) \cdot \hat{i} $

Similarly, $\vec{a} \cdot \hat{j}=a_2$ and $\vec{a} \cdot \hat{k}=a_3$ $\therefore(\vec{a} \cdot \hat{i}) \cdot \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \cdot \hat{k}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}=\vec{a}$

Hence, the value of the filler is $\vec{a}$.

Solution

If $|\vec{a}|=|\vec{b}|$ then $\vec{a}= \pm \vec{b}$ which is true.

Hence, the statement is True.

Solution

True

Solution

Given that $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$

Squaring both sides, we get

$ \begin{matrix} |\vec{a}+\vec{b}|^{2} =|\vec{a}-\vec{b}|^{2} \\ \Rightarrow \qquad |\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b} =|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b} \\ \Rightarrow \qquad 2 \vec{a} \cdot \vec{b} =-2 \vec{a} \cdot \vec{b} \Rightarrow \vec{a} \cdot \vec{b}=-\vec{a} \cdot \vec{b} \\ \Rightarrow \qquad 2 \vec{a} \cdot \vec{b} =0 \Rightarrow \vec{a} \cdot \vec{b}=0 \end{matrix} $

which implies that $\vec{a}$ and $\vec{b}$ are orthogonal.

Hence the given statement is True.

Solution

$ \begin{aligned} (\vec{a}+\vec{b})^{2} & =(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b}) \\ & =|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b} \end{aligned} $

Hence, the given statement is False.

Solution

If $\vec{a} \cdot \vec{b}=0$ then $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos 90^{\circ}$

So the angle between the adjacent sides of the rhombus should be $90^{\circ}$ which is not possible.

Hence, the given statement is False.