Question

Let $a=2\hat{i}+\hat{j}+\hat{k},b=\hat{i}+2\hat{j}-\hat{j}$ an a unit vector $\hat{c}$ be coplanar. If $\hat{c}$ is perpendicular to $a$, then $c$ is equal to

• A. $(-\hat{j}+\hat{k})$
• B. $\pm \frac{1}{\sqrt{2}}(-\hat{j}+\hat{k})$
• C. $\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$
• D. None of these

Answer 1

The correct option is B $\pm \frac{1}{\sqrt{2}}(-\hat{j}+\hat{k})$

vector $a\times b$ will be perpendicular to plane of vectors

$a$ and $b$.

Hence, the vector $a\times (a\times b)$ will lie in the plane of $a$ and $b$.

as per given condition $c$ is a unit vector in the plane of $a$ and $b$.

Thus $c=$ unit vector along $a\times (a\times b)$

Here, $a\times (a\times b)=(a\cdot b)a-(a\cdot a)b$
$=\left[\right(2\hat{i}+\hat{j}+\hat{k}\left)\right(\hat{i}+2\hat{j}-\hat{k}\left)\right](2\hat{i}+\hat{j}+\hat{k})-\left[|a{|}^2\right](\hat{i}+2\hat{j}-\hat{k})$
$=[2+2-1](2\hat{i}+\hat{j}+\hat{k})=-(\sqrt{2^2+1^2+1^2}{)}^2$
$=3(2\hat{i}+\hat{j}+\hat{k})-\left(6\right)(\hat{i}+2\hat{j}-\hat{k})$
$=-9\hat{j}+9\hat{k}$
$\therefore$ Required unit vector $=\pm \frac{a\times (a\times b)}{|a\times (a\times b)|}$
$=\frac{\pm 9(-\hat{j}+\hat{k})}{9\sqrt{1^2+1^2}}$
$=\pm \frac{(-\hat{j}+\hat{k})}{\sqrt{2}}$.