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

Equation of the plane perpendicular to the line x1=y1=z3 and passing through the point (2, 3, 4) is.

A
x+2y+3z=9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x+y+3z=17
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2x+3y+z=17
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3x+2y+z=16
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B x+y+3z=17
The given line is parallel to a vector whose direction ratios are 1,1,3
So if the plane is perpendicular to the given line, direction rations of normal of the plane will be 1,1,3
Hence equation of plane passing through (2,3,4) is given by
(x2)1+(y3)1+(z4)3=0
x+y+3z=12+3+2
x+y+3z=17

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Tango With Straight Lines !!
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon