Let a line L passing through the point P(1,1,1) be perpendicular to the lines Let the line L intersect the -plane at the point Q. Another line parallel to L and passing through the point S(1, 0, -1) intersects the -plane at the point R. [2026]

Question

Let a line L passing through the point P(1,1,1) be perpendicular to the lines $\frac{x - 4}{4} = \frac{y - 1}{1} = \frac{z - 1}{1} and \frac{x - 17}{1} = \frac{y - 71}{1} = \frac{z}{0} .$ Let the line L intersect the $y z$ -plane at the point Q. Another line parallel to L and passing through the point S(1, 0, -1) intersects the $y z$ -plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ________ . [2026]

Smart Achievers · Accepted Answer

(6)

$d_{1} = ⟨ 4, 1, 1 ⟩$ and $d_{2} = ⟨ 1, 1, 0 ⟩$

$d_{L} = d_{1} \times d_{2} =$ $| \begin{matrix} i & j & k \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{matrix} |$ $= ⟨ - 1, 1, 3 ⟩$

$Line L passes through P ⟨ 1, 1, 1 ⟩ with direction d_{L} = ⟨ - 1, 1, 3 ⟩$

$r (t) = ⟨ 1, 1, 1 ⟩ + t ⟨ - 1, 1, 3 ⟩ = ⟨ 1 - t, 1 + t, 1 + 3 t ⟩$

$For point Q, x = 0 \Rightarrow t = 1$

$Q = ⟨ 0, 2, 4 ⟩$

$Another line parallel to L passes through S ⟨ 1, 0, - 1 ⟩$

$r' (μ) = ⟨ 1, 0, - 1 ⟩ + μ ⟨ - 1, 1, 3 ⟩ = ⟨ 1 - μ, μ, - 1 + 3 μ ⟩$

$For point R, x = 0 \Rightarrow μ = 1$

$R = ⟨ 0, 1, 2 ⟩$

$Area of parallelogram with adjacent vectors \vec{P Q} and \vec{P S}$

$\vec{P Q} = ⟨ - 1, 1, 3 ⟩$

$\vec{P S} = ⟨ 0, - 1, - 2 ⟩$

Area of parallelogram

$\vec{P Q} \times \vec{P S} =$ $| \begin{matrix} i & j & k \\ - 1 & 1 & 3 \\ 0 & - 1 & - 2 \end{matrix} |$ $= ⟨ 1, - 2, 1 ⟩$

$Area = \sqrt{1^{2} + {(- 2)}^{2} + 1^{2}} = \sqrt{6}$

Solution

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