SATHEE: 3D Geometry 3 Question 55

3D Geometry 3 Question 55

55. Let $P$ be a point in the first octant, whose image $Q$ in the plane $x + y = 3$ (that is, the line segment $P Q$ is perpendicular to the plane $x + y = 3$ and the mid-point of $P Q$ lies in the plane $x + y = 3$ ) lies on the $Z$ -axis. Let the distance of $P$ from the $X$ -axis be 5 . If $R$ is the image of $P$ in the $X Y$ -plane, then the length of $P R$ is

(2018 Adv.)

Passage Based Problems

Read the following passage and answer the questions. Consider the lines

$L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}, L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$

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Answer:

Correct Answer: 55. (b)

Solution:

Let $P (α, β, γ)$ and $R$ is image of $P$ in the $x y$ -plane.

$∴ R (α, β, - γ)$

Also, $Q$ is the image of $P$ in the plane $x + y = 3$

$\begin{aligned} ∴ & \frac{x - α}{1} = \frac{y - β}{1} = \frac{z - γ}{0} = \frac{- 2 (α + β - 3)}{2} \\ x = 3 - β, y = 3 - α, z = γ \end{aligned}$

Since, $Q$ is lies on $Z$ -axis

$∴ β = 3, α = 3, z = γ$

$∴ P (3, 3, γ)$

Given, distance of $P$ from $X$ -axis be 5 .

$\begin{aligned} ∴ & 5 & = \sqrt{3^{2} + γ^{2}} \\ 25 - 9 & = γ^{2} \\ \Rightarrow & γ & = \pm 4 \\ Then, & P R & = | 2 γ | = | 2 \times 4 | = 8 \end{aligned}$

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3D Geometry 3 Question 55

Passage Based Problems

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

3D Geometry 3 Question 55

Passage Based Problems

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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