Consider the lines : x – 1 = y – 2 = z and : x – 2 = y = z – 1. Let the feet of the perpendiculars from the point P(5, 1, –3) on the lines and be Q and R respectively. If the area of the triangle PQR is A, then is equal to : [2025]

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Solution

Q.
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 $4 A^{2}$ is equal to : [2025]

1 147

2 143

3 139

4 151

Ans.
(1)

We have, the point P(5, 1, –3)

$L_{1}$ : x – 1 = y – 2 = z = $λ$ (say)

$L_{2}$ : x – 2 = y = z – 1 = $μ$ (say)

Any point on $L_{1}$ and $L_{2}$ are given by $Q (λ + 1, λ + 2, λ)$ and $R (μ + 2, μ, μ + 1)$ respectively.

$\vec{P Q} = (λ - 4) \hat{i} + (λ + 1) \hat{j} + (λ + 3) \hat{k}$

Since, $\vec{P Q} ⊥ L_{1}$ , whose direction ratios are < 1, 1, 1 >, So we have

$1 (λ - 4) + 1 (λ + 1) + 1 (λ + 3) = 0$

$\Rightarrow 3 λ = 0 \Rightarrow λ = 0$

$∴$ Q(1, 2, 0) is the foot of perpendicular on $L_{1}$ . Similarly,

$\vec{P R} = (μ - 3) \hat{i} + (μ - 1) \hat{j} + (μ + 4) \hat{k}$

Now, $\vec{P Q} ⊥ L_{2}$ , whose direction ratios are < 1, 1, 1 >, so we have

$1 (μ - 3) + 1 (μ - 1) + 1 (μ + 4) = 0$

$\Rightarrow 3 μ - 4 + 4 = 0 \Rightarrow μ = 0$

$∴$ R(2, 0, 1) is foot of perpendicular on $L_{2}$ .

Now, area $△ P Q R = \frac{1}{2}$ $| \vec{P Q} \times \vec{P R} |$

$\Rightarrow A = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 4 & 1 & 3 \\ - 3 & - 1 & 4 \end{matrix} | |$ $= \frac{1}{2}$ $| 7 \hat{i} + 7 \hat{j} + 7 \hat{k} |$ $= \frac{7 \sqrt{3}}{2}$

$∴ 4 A^{2} = 4 \times {(\frac{1}{2} \times 7 \sqrt{3})}^{2} = 147$

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