Let P be a point in the plane of the vectors A B =3 i + j - k and A C = i - j +3 k such that P is equidistant from the lines AB and AC. If | AP |= 5 2 , then the area of the triangle ABP is:

Step-by-step solution

Since P is equidistant from lines AB and AC through point A, it lies along the angle bisector direction. The unit vectors are u _ AB = 3 i + j - k 11 and u _ AC = i - j + 3 k 11 . The bisector direction is proportional to 2 i + k , so AP = (2 i + k ) . From | AP | = 5 2 : | | 5 = 5 2 , giving = 1 2 . Thus AP = i + 1 2 k . Area of triangle ABP = 1 2 | AB AP | . Computing: AB AP = 1 2 i - 5 2 j - k . | AB AP | = 1 4 + 25 4 + 1 = 30 4 = 30 2 . Area = 30 4 .