Find the vector equation of the plane which is at a distance of \frac{7}{\sqrt{38}} from the origin and the normal vector from origin is 2\hat{i}+3\hat{j}-5\hat{k}?

Category: QuestionsFind the vector equation of the plane which is at a distance of \frac{7}{\sqrt{38}} from the origin and the normal vector from origin is 2\hat{i}+3\hat{j}-5\hat{k}?
Editor">Editor Staff asked 11 months ago

Find the vector equation of the plane which is at a distance of \frac{7}{\sqrt{38}} from the origin and the normal vector from origin is 2\hat{i}+3\hat{j}-5\hat{k}?
 
(a) \vec{r}.(\frac{2\hat{i}}{38}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{56}}
 
(b) \vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}
 
(c) \vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{5\hat{j}}{\sqrt{38}}+\frac{3\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}
 
(d) \vec{r}.(\frac{2\hat{i}}{\sqrt{58}}-\frac{3\hat{j}}{\sqrt{37}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}
 
The question was asked in exam.
 
My question is from Three Dimensional Geometry topic in chapter 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

1 Answers
Editor">Editor Staff answered 11 months ago

The correct answer is (b) \vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}
 
The best I can explain: Let \vec{n}=2\hat{i}+3\hat{j}-5\hat{k}
 
\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{(2^2+3^2+(-5)^2)}}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{38}}
 
Hence, the required equation of the plane is \vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}