wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

The unit vector which is perpendicular to the vectors A=^i+2^j^k and B=3^i2^j+^k is

A
15(^j2^k)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(^j2^k)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
15(^j+4^k)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(^j+4^k)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 15(^j2^k)
We know that the unit vector perpendicular to two vectors is given as
^n=A×B|A×B|
So, we have A×B as per determinant method as
^i^j^k121321
(22)^i(1(3))^j+(26)^k
4^j8^k
Also, |A×B|=(4)2+(8)2=80=45
Hence, ^n=A×B|A×B|=4^j8^k45=15(^j2^k)

flag
Suggest Corrections
thumbs-up
38
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Basic Operations
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon