Question

If vectors `vecA=cos(omegat)hati +sin(omegat)hatj` and `vecB=cos((omegat)/2)hati+sin((omegat)/2)hatj` are functions of time, then the value of t at which they are orthogonal to each other is?

Answer 1

`t= pi/omega + (2npi)/omega; n in ZZ`

Explanation:

The vectors being orthogonal implies that the dot product is 0:

`vecA*vecB = 0`

One way to compute the dot product is to multiply the the two `hati` components and then add the product of the two `hatj` components:

`cos(omegat)cos((omegat)/2)+ sin(omegat)sin((omegat)/2)= 0`

Using the identity, `cos(A)cos(B) + sin(A)sin(B) = cos(A-B)`, we may substitute into the the above equation:

`cos(omegat-(omegat)/2)= 0`

Simplify:

`cos((omegat)/2)= 0`

Use the inverse cosine on both sides:

`(omegat)/2= cos^-1(0)`

Substitute the primary value `pi/2= cos^-1(0)`:

`(omegat)/2= pi/2`

Add the fact that this condition repeats at integer multiples of `pi`:

`(omegat)/2= pi/2 + npi; n in ZZ`

Multiply both sides of the equation by `2/omega`:

`t= pi/omega + (2npi)/omega; n in ZZ`