Find the equation of the plane passing through three points (1,2,-1), (0,-1,2) and (3,1,1).

Category: QuestionsFind the equation of the plane passing through three points (1,2,-1), (0,-1,2) and (3,1,1).
Editor">Editor Staff asked 11 months ago

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) \((\vec{r}-(\hat{i}+2\hat{j})).[(-\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0
 
I got this question in exam.
 
The origin of the question is Three Dimensional Geometry topic in portion Three Dimensional Geometry of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

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