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Find the equation of the plane passing through three points (1,2,-1), (0,-1,2) and (3,1,1).(a) \((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(-\hat{j}+2\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0(b) \((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0(c) \((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(3\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0(d) \((\vec{r}-(\hat{i}+2\hat{j})).[(-\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0I got this question in exam.The origin of the question is Three Dimensional Geometry topic in portion Three Dimensional Geometry of Mathematics – Class 12

Answer»

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

Explanation: LET \(\vec{a}=\hat{i}+2\hat{j}-\hat{k}, \,\vec{b}=-\hat{j}+2\hat{k}, \,\vec{c}=3\hat{i}+\hat{j}+\hat{k}\)

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

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

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


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