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

Q.45. Find a unit vector parallel to the xy plane and perpendicular to 4i^-3j^+k^.

Open in App
Solution

Dear student
Since the vector is parallel to the xy-plane, its z-component is zero. So the vector is of the form ai + bj. As it is perpendicular to 4i -3j +k, the dot product of the two is zero. This gives a condition on the values of a and b relative to one another. (ai + bj).(4i -3j +k) =0 4a - 3b = 0. So a =3b4. There are infinitely many such vectors, but you are after a unit vector. There will only be two of these. Being a unit vector, we know that a and b must satisfy a2+ b2= 1. Since a =3b4, we can substitute to find values of a and b to answer the question. 3b42+ b2 = 1 916+1b2 = 1 b2 = 1625. So b = 45 or b = -45 The corresponding a values are a = 35 or a =-35 This gives two solutions to choose from 35 i +45j and -35 i-45j You can easily double check to see that both of these satisfy all necessary requirements.
Regards

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Lines and Points
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon