wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A plane passes through (1,2,1) and is perpendicular to two planes 2x2y+z=0 and xy+2z=4. The distance of the plane from the point (1,2,2) is

A
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
22
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 22
The equation of a plane passing through (1,2,1) is
a(x1)+b(y+2)+c(z1)=0 .........(i)
It is perpendicular to 2x2y+z=0 and xy+2z=4.
Therefore, 2a2b+c=0 and ab+2c=0
a4+1=b41=c2+2
a3=b3=c0
On substituting the values of a,b and c in equation (i), we get
3(x1)3(y+2)+0(z1)=0
x+y+1=0
Distance of this plane from (1,2,2) is given by,
d=1+2+11+1=22

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Perpendicular Distance of a Point from a Plane
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon