SATHEE: 3D Geometry 3 Question 37

3D Geometry 3 Question 37

37. The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

(2015 Main)

(a) $2 \sqrt{14}$

(b) 8

(d) 13

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Answer:

Correct Answer: 37. (c)

Solution:

Given equation of line is

$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = λ$

and equation of plane is

$x - y + z = 16$

Any point on the line (i) is, $(3 λ + 2, 4 λ - 1, 12 λ + 2)$

Let this point of intersection of the line and plane.

$\begin{aligned} ∴ & (3 λ + 2) - (4 λ - 1) + (12 λ + 2) & = 16 \\ \Rightarrow & 11 λ + 5 = 16 \\ \Rightarrow & 11 λ = 11 \\ \Rightarrow & λ = 1 \end{aligned}$

So, the point of intersection is $(5, 3, 14)$ .

Now, distance between the points $(1, 0, 2)$ and $(5, 3, 14)$

$\begin{aligned} = \sqrt{(5 - 1)^{2} + (3 - 0)^{2} + (14 - 2)^{2}} \\ = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \end{aligned}$

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3D Geometry 3 Question 37

37. The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

Answer:

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3D Geometry 3 Question 37

37. The distance of the point (1,0,2) from the point of intersection of the line x−23=y+14=z−212 and the plane x−y+z=16 is

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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37. The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is