SATHEE: 3D Geometry 3 Question 32

3D Geometry 3 Question 32

####32. If the image of the point $P (1, - 2, 3)$ in the plane $2 x + 3 y - 4 z + 22 = 0$ measured parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is $Q$ , then $P Q$ is equal to

(a) $3 \sqrt{5}$

(b) $2 \sqrt{42}$

(d) $6 \sqrt{5}$

(2017 Main)

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Answer:

Correct Answer: 32. (b)

Solution:

Any line parallel to $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ and passing through $P (1, - 2, 3)$ is

$\frac{x - 1}{1} = \frac{y + 2}{4} = \frac{z - 3}{5} = λ$

Any point on above line can be written as

$(λ + 1, 4 λ - 2, 5 λ + 3)$

$∴$ Coordinates of $R$ are $(λ + 1, 4 λ - 2, 5 λ + 3)$ .

Since, point $R$ lies on the above plane.

$∴ 2 (λ + 1) + 3 (4 λ - 2) - 4 (5 λ + 3) + 22 = 0$

$\Rightarrow λ = 1$

So, point $R$ is $(2, 2, 8)$ .

Now, $P R = \sqrt{(2 - 1)^{2} + (2 + 2)^{2} + (8 - 3)^{2}} = \sqrt{42}$

$∴ P Q = 2 P R = 2 \sqrt{42}$

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3D Geometry 3 Question 32

Answer:

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