Question

Let A =i A cos theta +j A sin theta,be any vector.Another vector B which is normal to A is:
A) iB cos theta +j B sin theta
B)iB sin theta +j B cos theta
C)i B sin theta -j B cos theta
D)i A cos theta -j A sin theta

Answer 1

Final Answer : C)
p is theta angle
Steps:
1) A = cos(p) i + sin(p) j
We know, | A | = 1
Therefore,
Vector normal to A is given by
B = (k)dA/dp
= k [A (-sin(p)) i + A (cos (p)) j ]
where k is real number.

In options, k is -|B|/ | A¦
Therefore,
B = |B| sin(p) i - |B| cos(p) j

Answer 2

Explanation:

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