A = 10ar + 30aθ -10aφ    and B = -3ar - 10aθ +20aφ. Find a unit vector perpendicular to both vector A and vector Bans.0none of these-0.947ar + 0.322aθ + 0.019aφ0.947ar - 0.322aθ - 0.019aφ Previous Marked for Review Next

To find a unit vector that is perpendicular to both vectors A and B, we need to take the cross product of A and B. The cross product of two vectors is a vector that is perpendicular to both of the original vectors.

The cross product of A and B is given by:

C = A x B

The components of A are (10, 30, -10) and the components of B are (-3, -10, 20).

So,

C = (10, 30, -10) x (-3, -10, 20)

The cross product is calculated as follows:

Cx = AyBz - AzBy = 3020 - (-10)(-10) = 600 - 100 = 500 Cy = AzBx - AxBz = -10*(-3) - 1020 = 30 - 200 = -170 Cz = AxBy - AyBx = 10(-10) - 30*(-3) = -100 + 90 = -10

So, C = (500, -170, -10)

To convert C into a unit vector, we divide each component by the magnitude of C. The magnitude of C is given by:

|C| = sqrt(Cx^2 + Cy^2 + Cz^2) = sqrt(500^2 + (-170)^2 + (-10)^2) = sqrt(250000 + 28900 + 100) = sqrt(278900) = 528.3

So, the unit vector that is perpendicular to both A and B is:

C_unit = C / |C| = (500/528.3, -170/528.3, -10/528.3) = (0.947, -0.322, -0.019)

So, the answer is 0.947ar - 0.322aθ - 0.019aφ.