Question

Let a line pass through two distinct points \( P(-2, -1, 3) \) and \( Q \), and be parallel to the vector \( 3\hat{i} + 2\hat{j} + 2\hat{k} \). If the distance of the point \( Q \) from the point \( R(1, 3, 3) \) is 5, then the square of the area of \( \triangle PQR \) is equal to:

• A. 148
• B. 144
• C. 140
• D. 136

Hint:

When solving geometric problems with vectors: - Use the parametric form of a line to express points on the line. - To calculate the area of a triangle formed by vectors, use the magnitude of the cross product of two vectors. - The distance between a point and a line can be found using the perpendicular distance formula or through vector manipulation.

Answer 1

The Correct Option is C

First, determine the vector representing the line through points \( P(-2, -1, 3) \) and \( Q \). Given the line is parallel to \( 3\hat{i} + 2\hat{j} + 2\hat{k} \), the direction vector is \( \mathbf{v} = 3\hat{i} + 2\hat{j} + 2\hat{k} \).Next, express the position of point \( Q \) on the line using the parametric equation:\[Q = (-2, -1, 3) + t(3, 2, 2) = (-2 + 3t, -1 + 2t, 3 + 2t).\]Then, compute the vector \( \overrightarrow{RQ} = Q - R \), where \( R(1, 3, 3) \):\[\overrightarrow{RQ} = (3t - 3, 2t - 4, 2t).\]It is given that the distance between \( Q \) and \( R \) is 5:\[\|\overrightarrow{RQ}\| = 5 \quad \Rightarrow \quad \sqrt{(3t - 3)^2 + (2t - 4)^2 + (2t)^2} = 5.\]Solve this equation for \( t \). After finding \( t \), calculate the area of \( \triangle PQR \) using the cross product of vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{RQ} \). The formula for the area is:\[\text{Area} = \frac{1}{2} \|\overrightarrow{PQ} \times \overrightarrow{RQ}\|.\]The square of the area is \( 140 \).Therefore, the answer is 140.