Question

Consider the lines $ L_1: x - 1 = y - 2 = z $ and $ L_2: x - 2 = y = z - 1 $. Let the feet of the perpendiculars from the point $ P(5, 1, -3) $ on the lines $ L_1 $ and $ L_2 $ be $ Q $ and $ R $ respectively. If the area of the triangle $ PQR $ is $ A $, then $ 4A^2 $ is equal to:

• A. 139
• B. 147
• C. 151
• D. 143

Hint:

For calculating the area of a triangle in 3D, use the cross product of two vectors from the same point to the vertices, and then apply the formula for the area.

Answer 1

The Correct Option is B

We are given two lines and a point \( P(5, 1, -3) \), and we are required to find the area of the triangle \( PQR \), where \( Q \) and \( R \) are the feet of the perpendiculars from \( P \) onto the lines \( L_1 \) and \( L_2 \), respectively. 
The equations of the lines are given as: For \( L_1 \): \[ L_1: x - 1 = y - 2 = z. \] This can be written in parametric form as: \[ x = 1 + t, \quad y = 2 + t, \quad z = t. \] So, the direction ratios for \( L_1 \) are \( \mathbf{d_1} = (1, 1, 1) \). For \( L_2 \): \[ L_2: x - 2 = y = z - 1. \] This can be written in parametric form as: \[ x = 2 + s, \quad y = s, \quad z = 1 + s. \] So, the direction ratios for \( L_2 \) are \( \mathbf{d_2} = (1, 1, 1) \). 
Step 1: Find the Foot of the Perpendicular from \( P \) to \( L_1 \)
Let the foot of the perpendicular from \( P(5, 1, -3) \) to \( L_1 \) be \( Q \). The coordinates of \( Q \) will be of the form \( (1 + t, 2 + t, t) \). The vector \( \overrightarrow{PQ} \) is given by: \[ \overrightarrow{PQ} = (1 + t - 5, 2 + t - 1, t + 3) = (-4 + t, 1 + t, t + 3). \] Since \( \overrightarrow{PQ} \) is perpendicular to the direction vector \( \mathbf{d_1} = (1, 1, 1) \), we use the condition for perpendicularity, which is the dot product of \( \overrightarrow{PQ} \) and \( \mathbf{d_1} \) being zero: \[ (-4 + t) + (1 + t) + (t + 3) = 0 \quad \Rightarrow \quad 3t = 0 \quad \Rightarrow \quad t = 0. \] Thus, \( Q = (1, 2, 0) \). 
Step 2: Find the Foot of the Perpendicular from \( P \) to \( L_2 \)
The direction ratios of \( L_2 \) are \( \mathbf{d_2} = (1, 1, 1) \), and similarly, the vector \( \overrightarrow{PR} \) is perpendicular to \( L_2 \). The coordinates of \( R \) will be of the form \( (2 + s, s, 1 + s) \). The vector \( \overrightarrow{PR} = (2 + s - 5, s - 1, 1 + s + 3) = (-3 + s, s - 1, 4 + s) \) is perpendicular to \( \mathbf{d_2} \). The dot product condition gives: \[ (-3 + s) + (s - 1) + (4 + s) = 0 \quad \Rightarrow \quad 3s = 0 \quad \Rightarrow \quad s = 0. \] Thus, \( R = (2, 0, 1) \). 
Step 3: Calculate the Area of Triangle \( PQR \)
The area of the triangle \( PQR \) is given by the formula for the area of a triangle in 3D, using the cross product of the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \): \[ \text{Area} = \frac{1}{2} \left| \overrightarrow{PQ} \times \overrightarrow{PR} \right|. \] The vectors \( \overrightarrow{PQ} = (-4, 1, 3) \) and \( \overrightarrow{PR} = (-3, -1, 4) \). The cross product is: \[ \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} -4 & 1 & 3 -3 & -1 & 4 \end{vmatrix} = \hat{i}(1 \cdot 4 - 3 \cdot (-1)) - \hat{j}(-4 \cdot 4 - 3 \cdot (-3)) + \hat{k}(-4 \cdot (-1) - 1 \cdot (-3)). \] Simplifying: \[ \overrightarrow{PQ} \times \overrightarrow{PR} = \hat{i}(4 + 3) - \hat{j}(-16 + 9) + \hat{k}(4 + 3) = 7\hat{i} + 7\hat{j} + 7\hat{k}. \] Thus, the magnitude of the cross product is: \[ \left| \overrightarrow{PQ} \times \overrightarrow{PR} \right| = \sqrt{7^2 + 7^2 + 7^2} = \sqrt{147}. \] The area of the triangle is: \[ \text{Area} = \frac{1}{2} \times \sqrt{147} = \frac{\sqrt{147}}{2}. \] Finally, we need to compute \( 4A^2 \). Since \( A = \frac{\sqrt{147}}{2} \), we have: \[ A^2 = \frac{147}{4}, \quad 4A^2 = 147. \] Thus, the correct answer is \( 147 \), which corresponds to option (2).