Question

Which of the given set of planes are perpendicular to each other?

(a) \(\vec{r}.(2\hat{i}+2\hat{j}+\hat{k})\)=5 and \(\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5

(b) \(\vec{r}.(\hat{i}-2\hat{j}+\hat{k})\)=7 and \(\vec{r}.(\hat{i}+\hat{j}+2\hat{k})\)=2

(c) \(\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})\)=4 and \(\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5

(d) \(\vec{r}.(3\hat{i}-2\hat{j}+\hat{k})\)=2 and \(\vec{r}.(\hat{i}+2\hat{j}+8\hat{k})\)=8

This question was posed to me at a job interview.

This intriguing question comes from Three Dimensional Geometry topic in portion Three Dimensional Geometry of Mathematics – Class 12

Answer 1

Right option is (c) \(\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})\)=4 and \(\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5

The explanation is: Consider the set of planes \(\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})=4 \,and \,\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5

For the set of planes to be perpendicular \(\vec{n_1.}\vec{n_2}\)=0

In the above set of planes, \(\vec{n_1}=2\hat{i}-2\hat{j}+\hat{k}\) and \(\vec{n_2}=\hat{i}+2\hat{j}+2\hat{k}\)

∴\(\vec{n_1}.\vec{n_2}\)=2(1)-2(2)+1(2)=0

Hence, they are perpendicular.