⇒Vectors

Correct answer option is (D)

Let $\bar{\mathrm{r}}=x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}$ then

$\begin{array}{rl}& \bar{\mathrm{r}}\times \bar{\mathrm{a}}=\bar{\mathrm{b}}\times \bar{\mathrm{a}}\Rightarrow (\bar{\mathrm{r}}-\bar{\mathrm{b}})\times \bar{\mathrm{a}}=\bar{0}\\ & \left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ x& y-2& \mathrm{z}+1\\ 1& 1& 0\end{array}\right|=\bar{0}\\ & \Rightarrow (-\mathrm{z}-1)\hat{\mathrm{i}}-(-\mathrm{z}-1)\hat{\mathrm{j}}+(x-y+2)\hat{\mathrm{k}}=\bar{0}\\ & \Rightarrow \mathrm{z}=-1,x-y=-2\text{.... (i)}\end{array}$

Now, $\bar{\mathrm{r}}\times \bar{\mathrm{b}}=\bar{\mathrm{a}}\times \bar{\mathrm{b}}=(\bar{\mathrm{r}}-\bar{\mathrm{a}})\times \bar{\mathrm{b}}=\bar{0}$

$\begin{array}{rl}& \therefore \left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ x-1& y-1& z\\ 0& 2& -1\end{array}\right|=\bar{0}\\ & \Rightarrow (1-y-2\mathrm{z})\hat{\mathrm{i}}-(1-x)\hat{\mathrm{j}}+(2x-2)\hat{\mathrm{k}}=\bar{0}\\ & \Rightarrow 1-y-2\mathrm{z}=0,x=1\text{.... (ii)}\end{array}$

Solving (i) and (ii), we get

$\begin{array}{rl}& x=1,y=3,z=-1\\ & \therefore \bar{\mathrm{r}}=\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}}\\ & |\bar{\mathrm{r}}|=\sqrt{1+9+1}=\sqrt{11}\\ & \therefore \frac{\bar{\mathrm{r}}}{|\bar{\mathrm{r}}|}=\frac{\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{11}}\end{array}$

Correct answer option is (A)

$\begin{array}{rl}& \bar{\mathrm{a}}\times (\bar{\mathrm{b}}\times \bar{\mathrm{c}})=\frac{\sqrt{3}}{2}(\bar{\mathrm{b}}+\bar{\mathrm{c}})\\ & \Rightarrow (\bar{\mathrm{a}}\cdot \bar{\mathrm{c}})\bar{\mathrm{b}}-(\bar{\mathrm{a}}\cdot \bar{\mathrm{b}})\bar{\mathrm{c}}=\frac{\sqrt{3}}{2}\bar{\mathrm{b}}+\frac{\sqrt{3}}{2}\bar{\mathrm{c}}\end{array}$

On comparing, we get $\bar{\mathrm{a}}\cdot \bar{\mathrm{c}}=\frac{\sqrt{3}}{2}$ and $\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}=-\frac{\sqrt{3}}{2}$

$\begin{array}{rl}& \Rightarrow |\bar{\mathrm{a}}||\bar{\mathrm{b}}|\cos \theta =-\frac{\sqrt{3}}{2}\dots [\because |\bar{\mathrm{a}}|=|\bar{\mathrm{b}}|=1]\\ & \Rightarrow \cos \theta =-\frac{\sqrt{3}}{2}\\ & \Rightarrow \theta =\frac{5\pi }{6}\end{array}$

Correct answer option is (B)

Given that angle between $\bar{\mathrm{a}}\times \bar{\mathrm{b}}$ and $\bar{\mathrm{c}}\$is\$\frac{\pi }{4}$

$\therefore |(\overline{a}\times \overline{b})\times \overline{c}|=|(\overline{a}\times \overline{b})||\overline{c}|\sin \frac{\pi }{4}$ ..... (i)

Now, $\bar{\mathrm{a}}\times \bar{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 1& -2\\ 1& 1& 0\end{array}\right|$

$\begin{array}{rl}& =\hat{\mathrm{i}}(0+2)-\hat{\mathrm{j}}(0+2)+\hat{\mathrm{k}}(2-1)\\ & =2\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}\\ \therefore |\overrightarrow{\mathrm{a}}\times \bar{\mathrm{b}}|=& \sqrt{2^2+(-2{)}^2+1}=3\end{array}$

Given, $\bar{\mathrm{a}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}$

$a=\sqrt{2^2+1^2+(-2{)}^2}=3$

Given, $|\overline{c}-\overline{a}|=2\sqrt{2}$

Squaring on both sides, we get

$\begin{array}{rl}& |\overline{c}{|}^2+|\overline{a}{|}^2-2\overline{a}\cdot \overline{c}=8\\ & \Rightarrow |\overline{c}{|}^2+3^2-2|\overline{c}|=8...[\because \overline{a}.\overline{c}=|\overline{c}|]\end{array}$

$\begin{array}{rl}& \Rightarrow |\overline{c}{|}^2-2|\overline{c}|+1=0\\ & \Rightarrow (|\overline{c}|-1{)}^2=0\\ & \Rightarrow |\overline{c}|=1\end{array}$

From (i),

$\begin{array}{rl}|(\bar{\mathrm{a}}\times \bar{\mathrm{b}})\times \bar{\mathrm{c}}|& =|(\bar{\mathrm{a}}\times \bar{\mathrm{b}})||\bar{\mathrm{c}}|\cdot \sin \frac{\pi }{4}\\ & =3\times 1\times \frac{1}{\sqrt{2}}\\ & =\frac{3}{\sqrt{2}}\end{array}$

Correct answer option is (A)

Given: $(\bar{\mathrm{a}}\times \bar{\mathrm{b}})\times \bar{\mathrm{c}}=\frac{1}{3}|\bar{\mathrm{b}}||\bar{\mathrm{c}}|\overrightarrow{\mathrm{a}}$

We know that,

$(\bar{\mathrm{a}}\times \bar{\mathrm{b}})\times \bar{\mathrm{c}}=(\bar{\mathrm{a}}\cdot \bar{\mathrm{c}})\bar{\mathrm{b}}-(\bar{\mathrm{b}}\cdot \bar{\mathrm{c}})\bar{\mathrm{a}}$

On comparing, we get

$\begin{array}{rl}& \frac{1}{3}|\bar{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-\bar{\mathrm{b}}\cdot \bar{\mathrm{c}}\\ \Rightarrow & \frac{1}{3}|\bar{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-|\bar{\mathrm{b}}||\overrightarrow{\mathrm{c}}|\cos \theta \\ \Rightarrow & \cos \theta =\frac{-1}{3}\\ \Rightarrow & {\cos }^2\theta =\frac{1}{9}\\ & {\sin }^2\theta =1-{\cos }^2\theta \\ & =1-\frac{1}{9}\\ \therefore & {\sin }^2\theta =\frac{8}{9}\\ \therefore & \sin \theta =\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}\end{array}$

Correct answer option is (A)

Given $\bar{\mathrm{a}}\cdot \bar{\mathrm{r}}=3$

$\bar{\mathrm{a}}\times \bar{\mathrm{r}}=\bar{\mathrm{b}}$

Let $\bar{\mathrm{r}}=x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+z\hat{\mathrm{k}}$

$\begin{array}{rl}\bar{\mathrm{a}}\times \bar{\mathrm{r}}& =\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 1& 1\\ x& y& \mathrm{z}\end{array}\right|\\ & =(\mathrm{z}-y)\hat{\mathrm{i}}-\hat{\mathrm{j}}(\mathrm{z}-x)+\hat{\mathrm{k}}(y-x)\end{array}$

Given $\bar{\mathrm{a}}\times \bar{\mathrm{r}}=\bar{\mathrm{b}}$

$\therefore (\mathrm{z}-y)\hat{\mathrm{i}}-(\mathrm{z}-x)\hat{\mathrm{j}}+(y-x)\hat{\mathrm{k}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$

Comparing

$\begin{array}{rl}& \mathrm{z}-y=0\text{..... (i)}\\ & \mathrm{z}-x=-1\text{..... (ii)}\\ & y-x=-1\text{..... (iii)}\end{array}$

Also, $\bar{\mathrm{a}}\cdot \bar{\mathrm{r}}=3$

$\begin{array}{rl}& (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})(x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+\mathrm{zk})=3\\ & x+y+z=3\text{..... (iv)}\end{array}$

Solving equations (i), (ii), (iii) and (iv), we get

$\begin{array}{rl}& x=\frac{5}{3},y=\frac{2}{3},\mathrm{z}=\frac{2}{3}\\ \therefore \bar{\mathrm{r}}& =\frac{5}{3}\hat{\mathrm{i}}+\frac{2}{3}\hat{\mathrm{j}}+\frac{2}{3}\hat{\mathrm{k}}\end{array}$