Find the Cartesian equation of the plane passing through the point (3,2,-3) and the normal to the plane is 4\hat{i}-2\hat{j}+5\hat{k}?

Category: QuestionsFind the Cartesian equation of the plane passing through the point (3,2,-3) and the normal to the plane is 4\hat{i}-2\hat{j}+5\hat{k}?
Editor">Editor Staff asked 11 months ago

Find the Cartesian equation of the plane passing through the point (3,2,-3) and the normal to the plane is 4\hat{i}-2\hat{j}+5\hat{k}?
 
(a) 4x-2y+5z+7=0
 
(b) 3x-2y-3z+1=0
 
(c) 4x-y+5z+7=0
 
(d) 4x-2y-z+7=0
 
I got this question in an international level competition.
 
The query is from Three Dimensional Geometry topic in division 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

The correct option is (a) 4x-2y+5z+7=0
 
Explanation: The position vector of the point (3,2,-3) is \vec{a}=3\hat{i}+2\hat{j}-3\hat{k} and the normal vector \vec{N} perpendicular to the plane is \vec{N}=4\hat{i}-2\hat{j}+5\hat{k}
 
Therefore, the vector equation of the plane is given by (\vec{r}-\vec{a}).\vec{N}=0
 
Hence, (\vec{r}-(3\hat{i}+2\hat{j}-3\hat{k})).(4\hat{i}-2\hat{j}+5\hat{k})=0
 
4(x-3)-2(y-2)+5(z+3)=0
 
4x-2y+5z+7=0.