Question

Let $\overrightarrow{a}=2\hat{i}+\hat{j}-2\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}$. If $\overrightarrow{c}$ is a vector such that $\overrightarrow{a}\cdot \overrightarrow{c}=|\overrightarrow{c}|,|\overrightarrow{a}-\overrightarrow{c}|=2\sqrt{2}$ and the angle between $\overrightarrow{a}\times \overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi }{6}$, then

• A. $|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=\frac{3}{2}$
• B. $|\overrightarrow{c}|=1$
• C. $|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=\frac{2}{3}$
• D. $|\overrightarrow{c}|=4$

Answer 1

Answer: (A), (B)

The correct options are
A $|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=\frac{3}{2}$
B $|\overrightarrow{c}|=1$

$|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=|\overrightarrow{a}\times \overrightarrow{b}||\overrightarrow{c}|\sin \frac{\pi }{6}\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 1& 1& 0\end{array}\right|\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=2\hat{i}-2\hat{j}+\hat{k}\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|=3$

Using $|\overrightarrow{a}-\overrightarrow{c}|=2\sqrt{2}$
$|\overrightarrow{a}-\overrightarrow{c}{|}^2=8\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{c}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{c}=8\Rightarrow 9+|\overrightarrow{c}{|}^2-2|\overrightarrow{c}|=8\Rightarrow |\overrightarrow{c}|=1$

So,
$|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=|\overrightarrow{a}\times \overrightarrow{b}||\overrightarrow{c}|\sin \frac{\pi }{6}\Rightarrow |(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=3\times 1\times \frac{1}{2}\Rightarrow |(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|=\frac{3}{2}$