CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

Find the equation of the plane passing through the point (1,11) and perpendicular to the plnes x+2y+3z7=0 and 2x3y+4z=0.

Open in App
Solution

Let the DR's of the normal to the required plane be a,b,c.
Since the required plane passes through the point (1,1,1).
So, its equation is :
a(x1)+b(y1)+c(z+1)=0 ........(1)
Also, plane 1 is perpendicular to each of the given planes. Therefore,
a+2b+3c=0 .......(2)
and 2a3b+4c=0 .........(3)
On solving equation 2 and 3, we get,
a17=b2=c7=λ

a=17λ,b=2λ,c=7λ
On putting the values of a, b, c in equation 1, we get,
17λ(x1)+2λ(y1)+(7λ)(z+1)=0
λ[17x+2y7z26]=0
17x+2y7z=26

flag
Suggest Corrections
thumbs-up
0
mid-banner-image
mid-banner-image
similar_icon
Related Videos
thumbnail
lock
Line in Three Dimensional Space
MATHEMATICS
Watch in App