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

The vector equation of the plane through the point i+2jk and to the line of intersection of the plane r.(3ij+k)=1 and r.(i+4j2k)=2 is

A
r.(2i+7j13k)=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
r.(2i7j13k)=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
r.(2i+7j+13k)=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C r.(2i7j13k)=1
The line of inetrsection of the planes r.(3ij+k)=1 and r.(i+4j2k)=2 is common to both the planes.
Therefore, it is to normals to the two planes i.e. n1=3ij+k and n2=i+4j2k.
Hence, it is parallel to the vector n1×n2=2i+7j+13k.
Thus, we have to find the equation of the plane passing through a=i+2jk and normal to the vector n=n1×n2.
The equation of the required plane is
(ra).nr.n=a.nr.(2i+7j+13k)=(i+2j+k).(2i+7j+13k)r.(2i7j13k)=1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Equation of a Plane - Normal Form
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon