SATHEE: 3D Geometry 3 Question 47

3D Geometry 3 Question 47

####47. A plane passes through $(1, - 2, 1)$ and is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , then the distance of the plane from the point $(1, 2, 2)$ is

(a) 0

(b) 1

(d) $2 \sqrt{2}$

(2006, 3M)

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Answer:

Correct Answer: 47. (d)

Solution:

Let the equation of plane be

$a (x - 1) + b (y + 2) + c (z - 1) = 0$

which is perpendicular to $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ .

$\begin{aligned} \Rightarrow 2 a - 2 b + c = 0 and a - b + 2 c = 0 \\ \Rightarrow \frac{a}{- 3} = \frac{b}{- 3} = \frac{c}{0} \Rightarrow \frac{a}{1} = \frac{b}{1} = \frac{c}{0} . \end{aligned}$

So, the equation of plane is $x - 1 + y + 2 = 0$ or

$x + y + 1 = 0$

Its distance from the point $(1, 2, 2)$ is $\frac{| 1 + 2 + 1 |}{\sqrt{2}} = 2 \sqrt{2}$

3D Geometry 3 Question 47

Answer:

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3D Geometry 3 Question 47

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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