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

The vector representing the line through \( P(-2, -1, 3) \) and \( Q \) is \( \mathbf{v} = 3\hat{i} + 2\hat{j} + 2\hat{k} \), as the line is parallel to \( 3\hat{i} + 2\hat{j} + 2\hat{k} \). Point \( Q \) on the line can be expressed parametrically as \( Q = (-2, -1, 3) + t(3, 2, 2) = (-2 + 3t, -1 + 2t, 3 + 2t) \). The vector \( \overrightarrow{RQ} \) is \( Q - R \), where \( R(1, 3, 3) \): \( \overrightarrow{RQ} = (3t - 3, 2t - 4, 2t) \). Given that the distance between \( Q \) and \( R \) is 5, we have \( \|\overrightarrow{RQ}\| = 5 \), which leads to \( \sqrt{(3t - 3)^2 + (2t - 4)^2 + (2t)^2} = 5 \). Solving for \( t \) and then calculating the area of \( \triangle PQR \) using the cross product \( \overrightarrow{PQ} \times \overrightarrow{RQ} \), where \( \text{Area} = \frac{1}{2} \|\overrightarrow{PQ} \times \overrightarrow{RQ}\| \), yields an area whose square is 140. Therefore, the answer is 140.