Question

The direction cosines of the unit vector which is passing through the origin and is perpendicular to the plane $\overrightarrow{r}\cdot (-2\hat{i}+4\hat{j}+4\hat{k})+3=0$ is

• A. $\frac{2}{3},\frac{-4}{3},\frac{-4}{3}$
• B. $\frac{-2}{3},\frac{4}{3},\frac{4}{3}$
• C. $\frac{-1}{3},\frac{2}{3},\frac{2}{3}$
• D. $\frac{1}{3},\frac{-2}{3},\frac{-2}{3}$

Answer 1

The correct option is D $\frac{1}{3},\frac{-2}{3},\frac{-2}{3}$

$\overrightarrow{r}\cdot (-2\hat{i}+4\hat{j}+4\hat{k})+3=0$
$\Rightarrow \overrightarrow{r}\cdot (2\hat{i}-4\hat{j}-4\hat{k})=3\cdots \left(1\right)$
$\text{Now,}|2\hat{i}-4\hat{j}-4\hat{k}|=\sqrt{36}=6$
Dividing $\left(1\right)$ by $6,$ we get
$\overrightarrow{r}\cdot (\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k})=\frac{1}{2}$
which is the equation of the form in $\overrightarrow{r}\cdot \hat{n}=d$
$\text{Here,}\hat{n}=\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k}$ is perpendicular to the plane and passes through the origin.
Therefore, the direction cosines of $\hat{n}$ are $\frac{1}{3},\frac{-2}{3},\frac{-2}{3}.$