Find the equation of the plane passing through the three points (2,2,0), (1,2,1), (-1,2,-2).

(a) \((\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]\)=0

(b) \((\vec{r}-(3\hat{i}-2\hat{k})).[(-\hat{i}+\hat{k})×(2\hat{i}-2\hat{j})]\)=0

(c) \((\vec{r}+(2\hat{i}+2\hat{j})).[(-\hat{i}-\hat{k})×(-3\hat{i}-2\hat{k})]\)=0

(d) \((\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}-\hat{k})×(3\hat{i}+2\hat{k})]\)=0

I have been asked this question during an interview.

Origin of the question is Three Dimensional Geometry in portion Three Dimensional Geometry of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

Correct option is (a) (\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]=0

For explanation I would say: Let \vec{a}=2\hat{i}+2\hat{j}, \,\vec{b}=\hat{i}+2\hat{j}+\hat{k}, \,\vec{c}=-\hat{i}+2\hat{j}-2\hat{k}

The vector equation of the plane passing through three points is given by

(\vec{r}-\vec{a}).[(\vec{b}-\vec{a})×(\vec{c}-\vec{a})]=0

(\vec{r}-(2\hat{i}+2\hat{j})).[((\hat{i}+2\hat{j}+\hat{k})-(2\hat{i}+2\hat{j}))×((-\hat{i}+2\hat{j}-2\hat{k})–(2\hat{i}+2\hat{j}))]=0

(\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]=0.