Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Find the equation of the plane passing through the points and perpendicular to the plane <img alt="2x - y + z

Find the equation of the plane passing through the points (1, 0, - 2), (3, - 1, 0)and perpendicular to the plane 2x - y + z = 8. Also find the distance of the plane thus obtained from the origin.

 

 
 
 
 
 

Answers (1)

\\$The equation of the plane through (1,0,-2)$ \\ a(x-1)+by+c(z+2)=0\cdots(i)\\ $ It passes through (3,-1,0) $ \\ a(3-1)-b+2c=0 \\ 2a-b+2c = 0.....(ii) \\ $This plane is perpendicular to the plane $ \\ 2 x- y+ z=8 \\ 2a -b +c = 0 ......(iii)\\

\frac{a}{1} = \frac{b}{2} = \frac{c}{0} = k

{a}= {k}, {b}= {2k}, {c}= {0}

Put value of a,b,c in equation (i)

\\ {k}(x-1)+2k(y)+{0}k(z+2)= 0 \\ x -1+2y = 0 \\ x+2y = 1

$ The distance of the plane from the origin$ =\frac{\left|d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} \\ = \frac{\left|1\right|}{\sqrt{1^{2}+2^{2}}}\\ =\frac{1}{\sqrt{5}} 

 

 

Posted by

Safeer PP

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 board exam 2026 scheduled on March 3 postponed for Class 10, 12; revised date sheet out
anam-khan

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

Latest Question