Question

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be unit vectors such the $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{a}\cdot \overrightarrow{c}=0$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi }{6}$ if $\overrightarrow{a}=n(\overrightarrow{b}\times \overrightarrow{c})$ , Then value of $n$ is

• A. $\pm 1$
• B. $\pm 2$
• C. $\pm \sqrt{3}$
• D. $0$

Answer 1

The correct option is B $\pm 2$

Given $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors.

$\hat{a}.\hat{b}=0=\hat{a}.\hat{c}$

$\Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cosP=0=\left|\overrightarrow{a}\right|\left|\overrightarrow{c}\right|cosQ$

where P and Q are the angles between $\overrightarrow{a}$ and

$\overrightarrow{b}$ and $\overrightarrow{a}$ and $\overrightarrow{c}$ respectively

$\Rightarrow$ cos P = cos Q = 0

$\Rightarrow \angle P=\angle Q=\pi /2$

Now, $\hat{b}\times \hat{c}=\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|sin\left(\pi /6\right)\hat{n}$

$=\frac{1}{2}\hat{n}$

where $\hat{n}$ is the unit vecctor in the direction

of $\hat{b}\times \hat{c}$ i.e., in the direction of $\hat{a}$

$\Rightarrow \hat{a}$ is parallel to $\hat{n}$ or

$\hat{a}$ is anti-parallel to $\hat{n}$

$\therefore \hat{n}=\pm \hat{a}$

so, $\hat{b}\times \hat{c}=\frac{1}{2}(\pm \hat{a})$

$\Rightarrow \hat{a}=\pm 2(\hat{b}\times \hat{c})$ $\therefore n=\pm 2$