Let P be the foot of the perpendicular from the point on the line . Then the area of the right angled triangle PQR, when R is the point (3, –2, 1), is [2025]

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Solution

Q.
Let P be the foot of the perpendicular from the point Q(10, –3, –1) on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2}$ . Then the area of the right angled triangle PQR, when R is the point (3, –2, 1), is [2025]

1 $8 \sqrt{15}$

2 $\sqrt{30}$

3 $9 \sqrt{15}$

4 $3 \sqrt{30}$

Ans.
(4)

Let $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2} = λ$ then

Point $P = (7 λ + 3, - λ + 2, - 2 λ - 1)$

Now, Dr's of $Q P = (7 λ - 7, - λ + 5, - 2 λ)$

$∴ (7 λ - 7) (7) + (- λ + 5) (- 1) + (- 2 λ) (- 2) = 0$

$\Rightarrow 49 λ - 49 + λ - 5 + 4 λ = 0 \Rightarrow λ = 1$

$∴ P = (10, 1, - 3)$

$∴ \vec{P Q} = - 4 \hat{j} + 2 \hat{k}, \vec{P R} = - 7 \hat{i} - 3 \hat{j} + 4 \hat{k}$ [ $∵$ R = (3, –2, 1)]

$∴ Area of △ P Q R = \frac{1}{2}$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & - 4 & 2 \\ - 7 & - 3 & 4 \end{matrix} |$

$= \frac{1}{2}$ $| - 10 \hat{i} - 14 \hat{j} - 28 \hat{k} |$

$= \frac{1}{2} \sqrt{100 + 196 + 784} = 3 \sqrt{30} sq . units$ .

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