SATHEE: 3D Geometry 3 Question 1

3D Geometry 3 Question 1

####1. The distance of the point $(1, - 5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is

(a) $3 \sqrt{10}$

(b) $10 \sqrt{3}$

(d) $\frac{20}{3}$

(2016 Main)

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Answer:

(b)

Solution:

Equation of line passing through the point $(1, - 5, 9)$ and parallel to $x = y = z$ is

$\frac{x - 1}{1} = \frac{y + 5}{1} = \frac{z - 9}{1} = λ$

Thus, any point on this line is of the form $(λ + 1, λ - 5, λ + 9)$ .

Now, if $P (λ + 1, λ - 5, λ + 9)$ is the point of intersection of line and plane, then

$\begin{aligned} (λ + 1) - (λ - 5) + λ + 9 & = 5 \\ \Rightarrow & λ + 15 = 5 \Rightarrow λ & = - 10 \end{aligned}$

$∴$ Coordinates of point $P$ are $(- 9, - 15, - 1)$ .

Hence, required distance

$\begin{aligned} = \sqrt{(1 + 9)^{2} + (- 5 + 15)^{2} + (9 + 1)^{2}} \\ = \sqrt{10^{2} + 10^{2} + 10^{2}} \\ = 10 \sqrt{3} \end{aligned}$

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3D Geometry 3 Question 1

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