Question

The unit normal vector to the plane $3x+2y-2z=8\sqrt{17}$ is

• A. $\frac{1}{\sqrt{3}}\,(\hat{i}+\hat{j}-\hat{k})$
• B. $\frac{1}{\sqrt{17}}\,(3\hat{i}+2\hat{j}-2\hat{k})$
• C. $\frac{1}{\sqrt{13}}\,(3\hat{i}+2\hat{j}+2\hat{k})$
• D. $\frac{1}{\sqrt{11}}\,(3\hat{i}+\hat{j}+\hat{k})$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{17}}\,(3\hat{i}+2\hat{j}-2\hat{k})$

Answer 2

We have the given equation of plane
$3x+2y-2z=8\sqrt{17}$
$\Rightarrow$ $(x\hat{i}+y\hat{j}+z\hat{k}).(3\hat{i}+2\hat{j}-2\hat{k})=8\sqrt{17}$
Which is of the form
$\vec{r}.\,\,\vec{n}\,=\,d$
Where $\vec{n}$ is the normal vector to the plane. Thus, required unit normal vector.
$\vec{n}=\frac{|\vec{n}|}{|\vec{n}|}=\frac{3\hat{i}+2\hat{j}-2\hat{k}}{\sqrt{9+4+4}}=\frac{1}{\sqrt{17}}\,(3\hat{i}+2\hat{j}-2\hat{k})$