Question

A\((3,5,-4)\), B\((-5,-5,-2)\), C\((-1,1,2)\) are the vertices of a triangle. \(P\) is a point on \(AB\) and \(CP\) is an angle bisector of \(\triangle ABC\). \(G\) is the centroid of the triangle \(ABC\). If the point which divides \(GP\) in the ratio \(1:3\) is \((\alpha,\beta,\gamma)\), then \(\alpha+\beta-\gamma=\)

• A. \(0\)
• B. \(1\)
• C. \(5\)
• D. \(3\)

Hint:

Use the Angle Bisector Theorem to locate the point on the opposite side. Then apply the section formula and the centroid formula successively.

Answer 1

The Correct Option is B

Step 1: Find the coordinates of \(P\). Since \[ CP \] is the angle bisector, \[ \frac{AP}{PB} = \frac{AC}{BC}. \] Now, \[ AC = \sqrt{(-4)^2+(-4)^2+6^2} = 2\sqrt{17}, \] and \[ BC = \sqrt{4^2+6^2+4^2} = 2\sqrt{17}. \] Hence, \[ AP=PB, \] so \(P\) is the midpoint of \(AB\). Therefore, \[ P = \left( \frac{3-5}{2}, \frac{5-5}{2}, \frac{-4-2}{2} \right) = (-1,0,-3). \]

Step 2: Find the centroid. The centroid of \[ \triangle ABC \] is \[ G = \left( \frac{3-5-1}{3}, \frac{5-5+1}{3}, \frac{-4-2+2}{3} \right) = (-1,\tfrac13,-\tfrac43). \]

Step 3: Find the required point. The point dividing \[ GP \] internally in the ratio \[ 1:3 \] is \[ \left( \frac{3(-1)+(-1)}4, \frac{3\left(\frac13\right)+0}4, \frac{3\left(-\frac43\right)+(-3)}4 \right) = \left( -1,\frac14,-\frac74 \right). \] Thus, \[ \alpha+\beta-\gamma = -1+\frac14+\frac74 = 1. \] Therefore, \[ \boxed{1}. \] Thus, \[ \boxed{(B)} \] is the correct answer.