Question

A given unit vector is orthogonal to $5\hat{i}+2\hat{j}+6\hat{k}$ and coplanar with $\hat{i}-\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}+\hat{k}$ then the vector is

• A. $\frac{3\hat{j}-\hat{k}}{\sqrt{10}}$
• B. $\frac{6\hat{i}-5\hat{k}}{\sqrt{61}}$
• C. $\frac{2\hat{i}-5\hat{k}}{\sqrt{29}}$
• D. $\frac{2\hat{i}+\hat{j}-2\hat{k}}{3}$

Answer 1

The correct option is A $\frac{3\hat{j}-\hat{k}}{\sqrt{10}}$

Let unit vector be $\hat{u}=\frac{a\hat{i}+b\hat{j}+c\hat{k}}{|\overrightarrow{u}|}$.

Let $\overrightarrow{v}=5\hat{i}+2\hat{j}+6\hat{k}$ , $\overrightarrow{x}=\hat{i}-\hat{j}+\hat{k}$ and $\hat{y}=2\hat{i}+\hat{j}+\hat{k}$.

Now as $\hat{u}$ and $\overrightarrow{v}$ are orthogonal $\therefore \hat{u}.\overrightarrow{v}=0$

$\frac{(a\hat{i}+b\hat{j}+c\hat{k})}{|\overrightarrow{u}|}.(5\hat{i}+2\hat{j}+6\hat{k})=0$

$5a+2b+6c=0\to \left(1\right)$

$\hat{u}$ is coplanar with $\overrightarrow{x}$ and $\overrightarrow{y}$ means there scalar triple product will be zero.

$\to \hat{u}.(\overrightarrow{x}\times \overrightarrow{y})=0$

$=\frac{(a\hat{i}+b\hat{j}+c\hat{k})}{\overrightarrow{u}}.(\overrightarrow{x}\times \overrightarrow{y})=0$

$=(a\hat{i}+b\hat{j}+c\hat{k}).(\overrightarrow{x}\times \overrightarrow{y})=0$

$\left|\begin{array}{ccc}a& b& c\\ 1& -1& 1\\ 2& 1& 1\end{array}\right|=0$

$=a(-1\times 1-1\times 1)-b(1\times 1-2\times 1)+c(1\times 1-2\times (-1\left)\right)=0$

$-2a+b+3c=0\to \left(2\right)$

Subtract equation $\left(1\right)-2\times \left(2\right)$

$7a=0$

Put $a=0$ in $\left(2\right)$, we get $b+3c=0\to b=-3c$

$\therefore \hat{u}=\frac{0\hat{i}-3c\hat{j}+c\hat{k}}{\sqrt{0^2+(-3c{)}^2+c^2}}=\frac{-3\hat{j}+\hat{k}}{\sqrt{10}}$

Now, there is no option like this, so may be the the vector has opposite direction then what we have find out. So we reversed direction by just multiplying with $-1$. Because both are correct just directions are opposite.

$\hat{u}=\frac{3\hat{j}-\hat{k}}{\sqrt{10}}$

Hence, $\left(A\right)$