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)

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}) $