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

The unit vector perpendicular to each of the vectors 2ij+k and 3i+4jk is

A
3i+5j+11k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(3i+5j+11k)/155
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(3i+5j+11k)/(155)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
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 (3i+5j+11k)/(155)
We know that a×b is vector perpendicular to the plane of a and b
Therefore unit vector perpendicular to the plane of a and b =a×b|a×b|
Now a×b=(2ij+k)×(3i+4jk)
=∣ ∣ijk211341∣ ∣=i(14)+j(3+2)+k(8+3)=3i+5j+11k
unit vector parallel to a×b
=a×b|a×b|=3i+5j+11k(32+52+112)
=3i+5j+11k155

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Basic Operations
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon