Question

Let \(A(4,3,-2)\), \(B(0,-4,2)\), and \(C(-4,7,6)\) be the vertices of a triangle \(ABC\). If \(D(p,q,r)\) is the point of intersection of the bisector of angle \(A\) and the side \(BC\), then \(2p+q+r=\)

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(0\)

Hint:

Whenever a point is formed by the intersection of an angle bisector and the opposite side of a triangle, immediately think of the Angle Bisector Theorem: \[ \frac{BD}{DC}=\frac{AB}{AC}. \] After finding the ratio, use the section formula to obtain the coordinates.

Answer 1

The Correct Option is A

Concept: The most important result required in this problem is the Angle Bisector Theorem. According to this theorem, if the internal bisector of an angle of a triangle intersects the opposite side, then it divides that side in the ratio of the lengths of the adjacent sides. If \(AD\) bisects \(\angle A\) in \(\triangle ABC\), then \[ \frac{BD}{DC}=\frac{AB}{AC}. \] Once the ratio in which \(D\) divides the segment \(BC\) is known, the coordinates of \(D\) can be obtained using the section formula in three dimensions. This problem combines:
• Distance formula in three dimensions.
• Angle Bisector Theorem.
• Section formula in coordinate geometry.

Step 1: Find the length of \(AB\). Using the distance formula, \[ AB=\sqrt{(0-4)^2+(-4-3)^2+(2+2)^2} \] \[ =\sqrt{16+49+16} \] \[ =\sqrt{81}=9. \] Thus, \[ AB=9. \]

Step 2: Find the length of \(AC\). Again using the distance formula, \[ AC=\sqrt{(-4-4)^2+(7-3)^2+(6+2)^2} \] \[ =\sqrt{64+16+64} \] \[ =\sqrt{144}=12. \] Therefore, \[ AC=12. \]

Step 3: Apply the Angle Bisector Theorem. Since \(AD\) bisects \(\angle A\), \[ \frac{BD}{DC}=\frac{AB}{AC} =\frac{9}{12} =\frac{3}{4}. \] Hence point \(D\) divides the segment \(BC\) internally in the ratio \[ 3:4. \]

Step 4: Use the section formula. Let \[ B(0,-4,2), \qquad C(-4,7,6). \] Since \(D\) divides \(BC\) in the ratio \(3:4\), \[ D= \left( \frac{3(-4)+4(0)}{3+4}, \frac{3(7)+4(-4)}{3+4}, \frac{3(6)+4(2)}{3+4} \right). \] Therefore, \[ D= \left( -\frac{12}{7}, \frac{21-16}{7}, \frac{18+8}{7} \right) \] \[ = \left( -\frac{12}{7}, \frac{5}{7}, \frac{26}{7} \right). \] Thus, \[ p=-\frac{12}{7},\quad q=\frac{5}{7},\quad r=\frac{26}{7}. \]

Step 5: Compute \(2p+q+r\). \[ 2p+q+r = 2\left(-\frac{12}{7}\right) +\frac{5}{7} +\frac{26}{7}. \] \[ = -\frac{24}{7} +\frac{31}{7} \] \[ = \frac{7}{7} =1. \] Hence, \[ \boxed{2p+q+r=1}. \]