Question

If $\overrightarrow{a}=2\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}$and a unit vector $\stackrel{\rightharpoonup }{c}$is coplanar. If $\overrightarrow{c}$is perpendicular to$\overrightarrow{a}$, then $\overrightarrow{c}$is equal to?

• A. $\pm \frac{1}{\sqrt{2}}\left(-\hat{j}+\hat{k}\right)$
• B. $\pm \frac{1}{\sqrt{3}}\left(-\hat{i}-\hat{j}+\hat{k}\right)$
• C. $\pm \frac{1}{\sqrt{5}}\left(\hat{i}-2\hat{j}\right)$
• D. $\pm \frac{1}{\sqrt{3}}\left(\hat{i}-\hat{j}-\hat{k}\right)$

Answer 1

The correct option is A

$\pm \frac{1}{\sqrt{2}}\left(-\hat{j}+\hat{k}\right)$

Explanation for the correct option:

Step1. Given the condition in the question :

Given $\overrightarrow{c}$is coplanar with $\overrightarrow{a}$and$\overrightarrow{b}$.

So, $\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}$

$$\begin{gathered} \Rightarrow c=x\left(2\hat{i}+\hat{j}+\hat{k}\right)+y\left(\hat{i}+2\hat{j}-\hat{k}\right) \\ \Rightarrow c=\left(2x+y\right)\hat{i}+\left(x+2y\right)\hat{j}+(x-y)\hat{k} \end{gathered}$$

Since $\overrightarrow{\mathbf{a}}\mathbf{\perp }\overrightarrow{\mathbf{c}}\mathbf{\Rightarrow }\overrightarrow{\mathbf{a}}\mathbf{·}\overrightarrow{\mathbf{c}}\mathbf{=}\mathbf{0}$

$$\begin{gathered} \Rightarrow 2·\left(2x+y\right)+1·\left(x+2y\right)+1·(x-y)=0 \\ \Rightarrow y=-2x \end{gathered}$$

Thus, $\overrightarrow{c}=-3x\hat{j}+3x\hat{k}=3x\left(-\hat{j}+\hat{k}\right)$

Step 2. To find the vector $\overrightarrow{c}$:

Since $\overrightarrow{c}$is a unit vector.

$$\begin{gathered} \Rightarrow \left|c\right|=1 \\ \Rightarrow \sqrt{{\left(-3x\right)}^2+{\left(3x\right)}^2}=1 \\ \Rightarrow 18x^2=1 \\ \Rightarrow x^2=\frac{1}{18} \\ \Rightarrow x=\pm \frac{1}{3\sqrt{2}} \end{gathered}$$

Therefore,

$\overrightarrow{c}=\pm 3·\frac{1}{3\sqrt{2}}(-\hat{j}+\hat{k})=\pm \frac{1}{\sqrt{2}}(-\hat{j}+\hat{k})$

Hence, the correct option is A.