Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Find the vector equation of a plane, which is at a distance of 5 units from the origin and whose normal vector is <img alt="2\hat{i}-\hat{j}+2\hat{k}\cdot" src="https://entrancecorner.codecogs.com/gif.latex?2%5Chat%7Bi%7D-%5Chat%7Bj%7D&

Find the vector equation of a plane, which is at a distance of 5 units from
the origin and whose normal vector is 2\hat{i}-\hat{j}+2\hat{k}\cdot

 

 

 

 
 
 
 
 

Answers (1)

\vec{n}= 2\hat{i}-\hat{j}+2\hat{k}
we know that  \vec{n}=\frac{\vec{n}}{\left | \vec{n} \right |}= \frac{ 2\hat{i}-\hat{j}+2\hat{k}}{\sqrt{\left ( 2 \right )^{2}+\left ( -1 \right )^{2}+\left ( 2 \right )^{2}}}
                      = \frac{ 2\hat{i}-\hat{j}+2\hat{k}}{3}= \frac{2}{3}\hat{i}-\frac{1}{3}\hat{j}+\frac{2}{3}\hat{k}
r\cdot \vec{n}= p  where p is the distance from origin
\Rightarrow r\cdot \left [\frac{2}{3}\hat{i} -\frac{1}{3}\hat{j} +\frac{2}{3}\hat{k}\right ]= 5
\Rightarrow r \left [2\hat{i}-\hat{j}+2\hat{k}]= 15

Posted by

Ravindra Pindel

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question