Let be the image of the point Q(3, –3, 1) in the line and R be the point (2, 5, –1). If the area of the triangle PQR is and , then K is equal to : [2024]

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Solution

Q.
Let $P (α, β, γ)$ be the image of the point Q(3, –3, 1) in the line $\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{- 1}$ and R be the point (2, 5, –1). If the area of the triangle PQR is $λ$ and $λ^{2} = 14 K$ , then K is equal to : [2024]

1 81

2 72

3 36

4 18

Ans.
(1)

$L : \frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{- 1} = μ$ be the given line.

Any point on L is given by $M (μ, μ + 3, - μ + 1)$

Direction ratios of $Q M = (μ - 3, μ + 6, - μ)$

Direction ratios of L = (1, 1, –1)

Now, QM $⊥$ L

So, we have, $1 \cdot (μ - 3) + 1 \cdot (μ + 6) - 1 \cdot (- μ) = 0$

$\Rightarrow 3 μ + 3 = 0$

$\Rightarrow μ = - 1$

So, M(–1, 2, 2) is mid point of PQ.

$\Rightarrow \frac{α + 3}{2} = - 1, \frac{β - 3}{2} = 2, \frac{γ + 1}{2} = 2$

$\Rightarrow P (α, β, γ) \equiv (- 5, 7, 3)$

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

$\Rightarrow λ = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & - 10 & - 2 \\ - 1 & 8 & - 2 \end{matrix} | |$

$\Rightarrow λ = \frac{1}{2}$ $| 36 \hat{i} + 18 \hat{j} + 54 \hat{k} |$

$\Rightarrow λ^{2} = \frac{1}{4} (4536)$

$\Rightarrow λ^{2} = 1134 = 14 K \Rightarrow k = \frac{1134}{14} = 81$ .

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