Let A, B , and C be three points on a circle with centre O . If the position vectors of these points with respect to O satisfy the relation 3 OC = - OA + 2 OB , then the angle subtended by the minor arc AC at the centre O is

Step-by-step solution

Let the radius of the circle be R . Then | OA | = | OB | = | OC | = R . Given 3 OC = - OA + 2 OB . Squaring both sides (taking the dot product with itself), we get: 3 | OC |^2 = | OA |^2 + 4 | OB |^2 - 4 ( OA OB ) Substituting the magnitudes: 3 R^2 = R^2 + 4 R^2 - 4 R^2 ( AOB) 3 = 5 - 4 ( AOB) 4 ( AOB) = 2 ( AOB) = 1 2 Thus, AOB = 3 . Now, take the dot product of OA with both sides of the given equation: 3 ( OA OC ) = -| OA |^2 + 2 ( OA OB ) 3 R^2 ( AOC) = -R^2 + 2 R^2 ( AOB) Substitute ( AOB) = 1 2 : 3 ( AOC) = -1