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

The equation of the plane perpendicular to the yz− plane and passing through the point (1,−2,4) and (3,−4,5) is

A
y+2z=5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2y+z=5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
y+2z=6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2y+z=6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C y+2z=6
Plane is perpendicular to yz plane
So it have x-axis as parallel to it
Find a vector lying in the plane using two given point as-
5=(31)^i+(4(2)^j+(5A)^k
5=2^i2^j+^k
parallel vector, p=1.^i+0.^j+0.^k
So normal to the plane,
N=S×P
=∣ ∣ ∣^i^j^k221100∣ ∣ ∣
=^j+2^k
equation of plane using a point and normal,
0.(x1)+1.(y(2)+2.(z4)=0
y+2+2z8=0
y+2z=6


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Foot of Perpendicular, Image and Angle Bisector
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon