Question

Unit vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar. A unit vector $\overrightarrow{d}$ is perpendicular to them. If $(\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{c}\times \overrightarrow{d})=\frac{1}{6}\hat{i}-\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}$ and the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is ${30}^o$, then $\overrightarrow{c}$ is/are :

• A. $\left(\frac{2\hat{i}+\hat{j}-\hat{k}}{3}\right)$
• B. $\pm \left(\frac{-\hat{i}+2\hat{j}-2\hat{k}}{3}\right)$
• C. $\left(\frac{-2\hat{i}-2\hat{j}+\hat{k}}{3}\right)$
• D. $\pm \left(\frac{-\hat{i}-2\hat{j}+2\hat{k}}{3}\right)$

Answer 1

The correct option is B $\pm \left(\frac{-\hat{i}+2\hat{j}-2\hat{k}}{3}\right)$

$(\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{c}\times \overrightarrow{d})=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{d}\right]\overrightarrow{c}-\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]\overrightarrow{d}$

Since $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar so

$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$

$\therefore (\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{c}\times \overrightarrow{d})=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{d}\right]\overrightarrow{c}$

Now, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{d}\right]=(\overrightarrow{a}\times \overrightarrow{b}).\overrightarrow{d}=|\overrightarrow{a}\times \overrightarrow{b}|.|\overrightarrow{d}|.\cos 0^o[\because \overrightarrow{d}\text{is perpendicular to both}\overrightarrow{a}\&\overrightarrow{b}\text{so parallel to}\overrightarrow{a}\times \overrightarrow{b}]=|\overrightarrow{a}\times \overrightarrow{b}|.|\overrightarrow{d}|=|\overrightarrow{a}|.|\overrightarrow{b}|.\sin {30}^o.|\overrightarrow{d}|[\because \text{angle between}\overrightarrow{a}\&\overrightarrow{b}is{30}^o]=1.1.\frac{1}{2}.(\pm 1).1\left[\text{all are unit modulus}\right]=\frac{1}{2}.(\pm 1)=\pm \frac{1}{2}$

$\therefore (\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{c}\times \overrightarrow{d})=(\pm )\frac{1}{2}\overrightarrow{c}=\frac{1}{6}\hat{i}-\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}$

$\Rightarrow \overrightarrow{c}=\pm 2.(\frac{1}{6}\hat{i}-\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k})=\pm (\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k})=\pm \left(\frac{\hat{i}-2\hat{j}+2\hat{k}}{3}\right)$