find a vector length of 5m in the xy plane that is perpendicular to A=3i+6j+2k (hint use dot product)​

Any vector in the xy-plane has the form v = c1i + c2j + 0k

For v to be perpendicular to a, we need v · a = 0

So, 3c1 - 5c2 = 0

For example, choose c1 = 5 and c2 = 3.

Then v = 5i + 3j + 0k is perpendicular to a.

ll v ll = √[52 + 32 + 02] = √34

A unit vector in the xy-plane that is perpendicular to 3i - 5j + k is:

u = v / llvll = (5/√34)i + (3/√34)j + 0k