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).

• Option (a): This option is incorrect because it represents the original vector $\hat{i}-2 \hat{j}+2 \hat{k}$, which does not have a magnitude of 9. The magnitude of this vector is $\sqrt{9} = 3$.

• Option (b): This option is incorrect because $\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}$ is the unit vector in the direction of $\hat{i}-2 \hat{j}+2 \hat{k}$, which has a magnitude of 1, not 9.

• Option (d): This option is incorrect because $9(\hat{i}-2 \hat{j}+2 \hat{k})$ is a vector that has a magnitude of $9 \times 3 = 27$, not 9.

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).

• Option (a): $\frac{3 \vec{a}-2 \vec{b}}{2}$ is incorrect because it does not correctly apply the section formula for dividing the join of two vectors in the given ratio. The correct application of the section formula results in a different expression.

• Option (b): $\frac{7 \vec{a}-8 \vec{b}}{4}$ is incorrect because it incorrectly combines the vectors and coefficients. The correct combination of the vectors and coefficients, according to the section formula, does not yield this result.

• Option (c): $\frac{3 \vec{a}}{4}$ is incorrect because it omits the contribution of the vector $\vec{b}$ entirely. The correct position vector should include terms involving both $\vec{a}$ and $\vec{b}$, but in this case, the $\vec{b}$ term cancels out, leading to a different coefficient for $\vec{a}$.

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).

• Option (a) $-\hat{i}+12 \hat{j}+4 \hat{k}$ is incorrect because the calculation of the vector components is wrong. The correct calculation for the $i$ component is $-3 - 2 = -5$, not $-1$. The $j$ component should be $7 - 5 = 2$, not $12$. The $k$ component is correctly calculated as $4 - 0 = 4$.

• Option (b) $5 \hat{i}+2 \hat{j}-4 \hat{k}$ is incorrect because the signs and values of the components are wrong. The $i$ component should be $-5$, not $5$. The $j$ component is correctly calculated as $2$. The $k$ component should be $4$, not $-4$.

• Option (d) $\hat{i}+\hat{j}+\hat{k}$ is incorrect because it does not match the calculated vector components. The correct components are $-5 \hat{i}$, $2 \hat{j}$, and $4 \hat{k}$, not $1 \hat{i}$, $1 \hat{j}$, and $1 \hat{k}$.

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).

• Option (a) $\frac{\pi}{6}$ is incorrect because $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, not $\frac{1}{2}$. The given condition $\cos \theta = \frac{1}{2}$ does not match this angle.

• Option (c) $\frac{\pi}{2}$ is incorrect because $\cos(\frac{\pi}{2}) = 0$, not $\frac{1}{2}$. The given condition $\cos \theta = \frac{1}{2}$ does not match this angle.

• Option (d) $\frac{5 \pi}{2}$ is incorrect because it is not a standard angle within the primary range of $0$ to $2\pi$ for which cosine values are typically considered. Additionally, $\cos(\frac{5 \pi}{2}) = 0$, not $\frac{1}{2}$. The given condition $\cos \theta = \frac{1}{2}$ does not match this angle.

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).

• Option (a) 0: If $\lambda = 0$, then the dot product of $\vec{a}$ and $\vec{b}$ would be $2 + 0 + 3 = 5$, which is not equal to 0. Therefore, $\vec{a}$ and $\vec{b}$ would not be orthogonal.

• Option (b) 1: If $\lambda = 1$, then the dot product of $\vec{a}$ and $\vec{b}$ would be $2 + 2 \cdot 1 + 3 = 7$, which is not equal to 0. Therefore, $\vec{a}$ and $\vec{b}$ would not be orthogonal.

• Option (c) $\frac{3}{2}$: If $\lambda = \frac{3}{2}$, then the dot product of $\vec{a}$ and $\vec{b}$ would be $2 + 2 \cdot \frac{3}{2} + 3 = 8$, which is not equal to 0. Therefore, $\vec{a}$ and $\vec{b}$ would not be orthogonal.

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).

• Option (b) $\frac{3}{2}$ is incorrect because if $\lambda = \frac{3}{2}$, the ratio $\frac{1}{\lambda}$ would be $\frac{2}{3}$, which does not match the required ratio $\frac{3}{2}$ for the vectors to be parallel.

• Option (c) $\frac{5}{2}$ is incorrect because if $\lambda = \frac{5}{2}$, the ratio $\frac{1}{\lambda}$ would be $\frac{2}{5}$, which does not match the required ratio $\frac{3}{2}$ for the vectors to be parallel.

• Option (d) $\frac{2}{5}$ is incorrect because if $\lambda = \frac{2}{5}$, the ratio $\frac{1}{\lambda}$ would be $\frac{5}{2}$, which does not match the required ratio $\frac{3}{2}$ for the vectors to be parallel.

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 $(d)$.

• Option (a) 340: This option is incorrect because the area of the triangle is calculated using the magnitude of the cross product of the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$, which results in $\frac{1}{2} \sqrt{229}$. The value 340 is not related to this calculation and is significantly larger than the correct value.

• Option (b) $\sqrt{25}$: This option is incorrect because $\sqrt{25}$ simplifies to 5, which is not the correct magnitude of the cross product of the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$. The correct magnitude is $\sqrt{229}$, and thus $\sqrt{25}$ is not the correct area.

• Option (c) $\sqrt{229}$: This option is incorrect because it represents the full magnitude of the cross product of the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$. The area of the triangle is half of this magnitude, so the correct answer is $\frac{1}{2} \sqrt{229}$, not $\sqrt{229}$.

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).

• Option (a) ( \vec{a}^2 ): This option is incorrect because the sum of the squares of the magnitudes of the cross products ((\vec{a} \times \hat{i})^2 + (\vec{a} \times \hat{j})^2 + (\vec{a} \times \hat{k})^2) results in twice the magnitude squared of (\vec{a}), not just the magnitude squared of (\vec{a}). The correct result is (2 \vec{a}^2), not (\vec{a}^2).

• Option (b) ( 3 \vec{a}^2 ): This option is incorrect because the sum of the squares of the magnitudes of the cross products does not result in three times the magnitude squared of (\vec{a}). The correct result is (2 \vec{a}^2), not (3 \vec{a}^2).

• Option (c) ( 4 \vec{a}^2 ): This option is incorrect because the sum of the squares of the magnitudes of the cross products does not result in four times the magnitude squared of (\vec{a}). The correct result is (2 \vec{a}^2), not (4 \vec{a}^2).

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 $(d)$.

• Option (a) 5 is incorrect because the calculation of ( |\vec{a} \times \vec{b}| ) using the given magnitudes and the sine of the angle between the vectors results in 16, not 5. The formula ( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta ) yields 16 when ( |\vec{a}| = 10 ), ( |\vec{b}| = 2 ), and ( \sin \theta = \frac{4}{5} ).

• Option (b) 10 is incorrect because, similar to option (a), the correct calculation using the given values results in 16. The product of the magnitudes and the sine of the angle between the vectors does not equal 10.

• Option (c) 14 is incorrect because the correct value of ( |\vec{a} \times \vec{b}| ) is 16, as derived from the given magnitudes and the sine of the angle between the vectors. The calculation does not yield 14.

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 $(a)$.

• Option (b) $\lambda=0$: This value does not satisfy the equation $\lambda^3 - 6\lambda - 4 = 0$. Substituting $\lambda = 0$ into the equation results in $-4 \neq 0$, hence it is incorrect.

• Option (c) $\lambda=1$: This value does not satisfy the equation $\lambda^3 - 6\lambda - 4 = 0$. Substituting $\lambda = 1$ into the equation results in $1 - 6 - 4 = -9 \neq 0$, hence it is incorrect.

• Option (d) $\lambda=-1$: This value does not satisfy the equation $\lambda^3 - 6\lambda - 4 = 0$. Substituting $\lambda = -1$ into the equation results in $-1 + 6 - 4 = 1 \neq 0$, hence it is incorrect.

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).

• Option (a) 1: This option is incorrect because the calculation shows that the sum of the dot products of the vectors is negative. Specifically, the sum of the dot products is (-\frac{3}{2}), not 1. The vectors (\vec{a}, \vec{b}, \vec{c}) being unit vectors and summing to zero implies that their pairwise dot products must sum to a negative value to balance the equation.

• Option (b) 3: This option is incorrect because the sum of the dot products of the vectors is (-\frac{3}{2}), not 3. The calculation clearly shows that the sum of the dot products is negative, and a value of 3 would not satisfy the given conditions of the problem.

• Option (d) None of these: This option is incorrect because the correct value of (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) is (-\frac{3}{2}), which is provided in option (c). Therefore, “None of these” is not the correct answer.

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).

• Option (b): $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ is incorrect because it represents a scalar quantity, not a vector. The projection vector should be a vector, not just a scalar.

• Option (c): $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$ is incorrect because it uses the magnitude of $\vec{a}$ in the denominator instead of the magnitude of $\vec{b}$. The projection of $\vec{a}$ on $\vec{b}$ should involve the magnitude of $\vec{b}$.

• Option (d): $(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^{2}}) \vec{b}$ is incorrect because it uses $|\vec{a}|^2$ in the denominator instead of $|\vec{b}|^2$. The correct formula for the projection vector involves the magnitude of $\vec{b}$ squared in the denominator.

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).

• Option (a) 0: This option is incorrect because the calculation shows that the sum of the dot products is not zero. The derived equation (38 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0) leads to (2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -38), which simplifies to (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -19), not 0.

• Option (b) 1: This option is incorrect because the derived equation (38 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0) simplifies to (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -19), not 1.

• Option (d) 38: This option is incorrect because the derived equation (38 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0) simplifies to (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -19), not 38.

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).

• Option (a) $[0,8]$ is incorrect because the range of $|\lambda \vec{a}|$ includes negative values, specifically from $-12$ to $8$, not just non-negative values.

• Option (c) $[0,12]$ is incorrect because the maximum value of $|\lambda \vec{a}|$ is $8$ when $\lambda = 2$, not $12$. Additionally, the range includes negative values down to $-12$.

• Option (d) $[8,12]$ is incorrect because the range of $|\lambda \vec{a}|$ starts from $-12$ and goes up to $8$, not starting from $8$.

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 $(b)$.

• Option (a) one: This option is incorrect because there are two distinct vectors of unit length that are perpendicular to both $\vec{a}$ and $\vec{b}$. These vectors are $\pm(\vec{a} \times \vec{b})$. Therefore, the number of such vectors is not one.

• Option (c) three: This option is incorrect because the cross product of two vectors $\vec{a}$ and $\vec{b}$ results in a single vector, and its negative, both of which are perpendicular to $\vec{a}$ and $\vec{b}$. There are no additional vectors that satisfy the condition, so the number of such vectors is not three.

• Option (d) infinite: This option is incorrect because the cross product $\vec{a} \times \vec{b}$ and its negative are the only two vectors of unit length that are perpendicular to both $\vec{a}$ and $\vec{b}$. There are not an infinite number of such vectors.

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.