Question

Let P be the foot of the perpendicular from the point \( Q(10,-3,-1) \) on the line: \[ \frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2}. \] Then the area of the right-angled triangle PQR, where R is the point \( (3,-2,1) \), is:

• A. \( 9\sqrt{15} \)
• B. \( \sqrt{30} \)
• C. \( 8\sqrt{15} \)
• D. \( 3\sqrt{30} \)

Hint:

When finding the area of a triangle using vectors, compute the determinant of a 3×3 matrix formed by the two vectors.

Answer 1

The Correct Option is D

Step 1: Parametrize the line. We have the equation of the line: \[ \frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z + 1}{-2} = \lambda. \] Thus, the parametric equations for the points on the line are: \[ x = 7\lambda + 3, \quad y = -\lambda + 2, \quad z = -2\lambda - 1. \] 

Step 2: Finding the foot of the perpendicular. The direction ratios of \( QP \) are given by: \[ 7\lambda - 7, \quad -\lambda + 5, \quad -2\lambda. \] We now solve for the value of \( \lambda \) such that the point \( P \) is the foot of the perpendicular from \( Q \). This leads to the equation: \[ (7\lambda - 7) \cdot 7 + (-\lambda + 5) \cdot (-1) + (-2\lambda) \cdot (-2) = 0. \] Solving this gives: \[ 54\lambda - 54 = 0 \quad \Rightarrow \quad \lambda = 1. \] Thus, the coordinates of point \( P \) are \( (10, 1, -3) \). 

Step 3: Finding vectors \( PQ \) and \( PR \). The coordinates of \( Q \) are \( (10, -3, -1) \), so the vector \( PQ \) is: \[ PQ = (10 - 10, 1 - (-3), -3 - (-1)) = 4j + 2k. \] The coordinates of \( R \) are \( (3, -2, 1) \), so the vector \( PR \) is: \[ PR = (10 - 3, 1 - (-2), -3 - 1) = -7i - 3j + 4k. \] 

Step 4: Finding the area of triangle PQR. The area of triangle PQR is given by the formula: \[ \text{Area} = \frac{1}{2} \left| \mathbf{PQ} \times \mathbf{PR} \right|. \] We now compute the cross product of vectors \( PQ \) and \( PR \): \[ \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} i & j & k 
0 & -4 & 2 
-7 & -3 & 4 \end{array} \right|. \] Expanding the determinant: \[ \text{Area} = \frac{1}{2} \left| (0)(-3 \cdot 4 - 2 \cdot 4) - (4)(-7 \cdot 4) + (2)(-7 \cdot -3) \right| \] \[ = \frac{1}{2} \times 30 = 3\sqrt{30}. \]