Question

let vector A is equals to 1 cos theta + sin theta by any vector another vector B which is normal to a is ​

Answer 1

Answer:

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