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. (d)

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 $(5, 3, 14)$

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

3D Geometry 3 Question 37

Answer:

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3D Geometry 3 Question 37

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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