Let the area of the triangle formed by the lines x + 2 = y – 1 = z, and be A. Then is equal to __________. [2025]

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Solution

Q.
Let the area of the triangle formed by the lines x + 2 = y – 1 = z, $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1}$ be A. Then $A^{2}$ is equal to __________. [2025]

Ans.
(56)

$L_{1} : x + 2 = y - 1 = z = λ (say)$

$L_{2} : \frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1} = μ (say)$

$L_{3} : \frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1} = δ (say)$

Any point of line $L_{1}, L_{2} and L_{3}$ is given by $(λ - 2, λ + 1, λ)$ , $(5 μ + 3, - μ, μ + 1)$ and $(- 3 δ, 3 δ + 3, δ + 2)$ respectively.

Point of intersection of $L_{1}$ and $L_{2}$ is given by

$\begin{matrix} λ - 2 = 5 μ + 3 \\ λ + 1 = - μ \\ λ = μ + 1 \end{matrix}} \Rightarrow λ = 0, μ = - 1$

$∴$ Point of intersection is A(–2, 1, 0).

Point of intersection of $L_{2}$ and $L_{3}$ is given by

$\begin{matrix} 5 μ + 3 = - 3 δ \\ - μ = 3 δ + 3 \\ μ + 1 = δ + 2 \end{matrix}} \Rightarrow μ = 0, δ = - 1$

$∴$ Point of intersection is B(3, 0 , 1).

Point of intersection of $L_{1}$ and $L_{3}$ is given by

$\begin{matrix} λ - 2 = - 3 δ \\ λ + 1 = 3 δ + 3 \\ λ = δ + 2 \end{matrix}} \Rightarrow λ = 2, δ = 0$

$∴$ Point of intersection is C(0, 3, 2).

Now, Area of $△ A B C = \frac{1}{2}$ $| \vec{A B} \times \vec{B C} |$

$∴ A = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 5 & 1 & - 1 \\ - 3 & 3 & 1 \end{matrix} | |$

$= \frac{1}{2} | 4 \hat{i} + 8 \hat{j} - 12 \hat{k |}$

$= \frac{1}{2} \sqrt{16 + 64 + 144} = \sqrt{56}$

$∴ A^{2} = 56$ .

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