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

The line joining the points (1,1,2) and (3,−2,1) meets the plane 3x+2y+z=6 at the point.

A
(1,1,2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(3,2,1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(2,3,1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(3,2,1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C (3,2,1)
The straight line joining the points (1,1,2) and (3,2,1) is x12=y13=z21=r(say)
Therefore, point is (2r+1,13r,2r) which lies on 3x+2y+z=6.
Thus 3(2r+1)+2(13r)+2r=6
r=1
So, required point is (3,2,1).

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Line in Three Dimensional Space
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon