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

Equation of the plane passing through the intersection of the planes x+y+z=6, 2x+3y+4z+5=0 and the point (1,1,1) is


A

20x+23y+26z69=0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

31x+45y+49z+52=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

8x+5y+2z69=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

4x+5y+6z7=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

20x+23y+26z69=0


Explanation for the correct option:

Given, equation of planes are

x+y+z=6

2x+3y+4z+5=0

Equation of plane passing through intersection of given plane is,

L1+λL2=0

x+y+z6+λ(2x+3y+4z+5)=0

it also passes through (1,1,1)

1+1+1+6+λ(2+3+4+5)=0

λ=314
the required plane is,

14(x+y+z6)+3(2x+3y+4z+5)=0

20x+23y+26z69=0

Hence, Option ‘A’ is Correct.


flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon