Question

The cartesian equation of the plane which is at a distance of 10 unit from the original and perpendicular to the vector i + 2j -2k is

• A. $x+2y+2z=30$
• B. $x-2y-2z=30$
• C. $x-2y+2z=30$
• D. $x+2y-2z=30$

Answer 1

The correct option is D $x+2y-2z=30$

Let the Normal vector $\overrightarrow{n}=\hat{i}+2\hat{j}-2\hat{k}$
$\therefore$Unit vector in the direction of $\overrightarrow{n}$

$\begin{array}{cc}& =\frac{\overrightarrow{n}}{|\overrightarrow{n}|}\\ & =\frac{\hat{i}+2\hat{j}-2\hat{k}}{\sqrt{1^2+2^2+(-2{)}^2}}\\ & =\frac{\hat{i}+2\hat{j}-2\hat{k}}{3}\\ & =\hat{n}\end{array}$

We know that position vector $\overrightarrow{r}$ and equation of plane are related as
$\overrightarrow{r}\cdot \hat{n}=d$
$\Rightarrow \overrightarrow{R}\cdot \frac{\hat{i}+2\hat{j}-2\hat{k}}{3}=10$

$\Rightarrow \overrightarrow{R}\cdot \hat{i}+2\hat{j}-2\hat{k}=30$
Required Equation of Plane is :
$x+2y-2z=30$