The resultant of two vectors P""⃗ and Q""⃗ is perpendicular to vector Q""⃗ . If the ratio of their magnitudes 01 = 2$ , the anglebetween the vectors is

The problem statement seems to be incomplete. However, I can provide a general solution based on the information given.

Given that the resultant of two vectors P and Q is perpendicular to vector Q, we can use the property of vectors that the square of the magnitude of the resultant vector (R) is equal to the sum of the squares of the magnitudes of the individual vectors (P and Q) plus 2 times the product of their magnitudes and the cosine of the angle between them (θ).

So, R² = P² + Q² + 2PQcos(θ)

But since R is perpendicular to Q, the dot product of R and Q will be zero. That is,

R.Q = |R||Q|cos(90°) = 0

This implies that,

P.Q + Q² = 0

or, P.Q = - Q² -----(1)

Also given that the ratio of their magnitudes P:Q = 1:2, we can substitute P = 2Q in equation (1) to get,

2Q.Q = - Q²

Solving this we get Q = 0 or Q = -2Q which implies that Q = 0.

So, the angle between the vectors P and Q would be 90 degrees or π/2 radians.

Please provide the complete problem statement for a more accurate solution.