Let a line pass through two distinct pointsP(–2, –1, 3) and Q, and be parallel to the vector . If the distance of the point Q from the point R(1, 3, 3) is 5, then the square of the area of PQR is equal to : [2025]

Question

Let a line pass through two distinct pointsP(–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 $△$ PQR is equal to : [2025]

Smart Achievers · Accepted Answer

(1)

Equation of line passing through P(–2, –1, 3) and parallel to vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ is $\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}$ .

Let $\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} = λ$

$\Rightarrow Q = (3 λ - 2, 2 λ - 1, 2 λ + 3), λ \in ℝ - {0}$

Also, $| \vec{Q R} |$ $= 5 = \sqrt{{(3 λ - 3)}^{2} + {(2 λ - 4)}^{2} + {(2 λ)}^{2}}$

$∴ 17 λ^{2} - 34 λ + 25 = 25 \Rightarrow λ = 2 (∵ λ \neq 0)$

Q(4, 3, 7), P(–2, –1, 3), R(1, 3, 3)

$∴ Area of △ P Q R = \frac{1}{2}$ $| \vec{P Q} \times \vec{P R} |$

$△ = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \end{matrix} | |$

$△ =$ $| - 8 \hat{i} + 6 \hat{j} + 6 \hat{k} |$ $= \sqrt{136} ∴ △^{2} = 136$

Solution

Q.
Let a line pass through two distinct pointsP(–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 $△$ PQR is equal to : [2025]

1 136

2 144

3 148

4 140

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