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Question

The distance of the point 3,8,2 from the line x-12=y-34=z-23 measured parallel to the plane 3x+2y-2z=0 is


A

2

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B

3

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C

6

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D

7

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Solution

The correct option is D

7


Explanation for the correct answer:

Step 1: Find the equation of the parallel plane to 3x+2y-2z=0 and passing through 3,8,2.

Let Equation of the plane parallel to given plane be 3x+2y-2z+c=0

This plane passes through 3,8,2

Therefore

3(3)+2(8)-2(2)+c=09+16-4+c=0c=-21

So equation of the plane is 3x+2y-2z-21=0

Step 2: Finding the coordinates of the point of intersection of plane and line.

We now find the intersection of this plane and the given line

let x-12=y-34=z-23=k

therefore,

x=2k+1y=4k+3z=3k+2

These points should satisfy equation of the plane, hence

32k+1+24k+3-23k+2-21=06k+3+8k+6-6k-4-21=08k-16=0k=2

So the coordinates of the point are x=5,y=11,z=8

Step 3: Finding the distance.

So the distance between 5,11,8and3,8,2 is given by:

(a1-a2)2+(b1-b2)2+(c1-c2)2

=5-32+11-82+8-22=4+9+36=49=7

Hence, option (D) is the correct answer.


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