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Question

Find the distance between the planes r(2i^-j^+3k^)-4=0 and r(6i^-3j^+9k^)+13=0.


A

5314

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B

10314

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C

25314

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D

None of these

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Solution

The correct option is C

25314


Explanation for the correct option:

Step 1. Simplify those two equations

Given equations are

r(2i^-j^+3k^)-4=0…..(i)

r(6i^-3j^+9k^)+13=0…..(ii)

Equation (ii) can be written as

r(2i^-j^+3k^)+133=0 …..(iii)

Step:2 Calculate the distance between two planes.

Distance between two lines are given bya1-a2x2+y2+z2,

We have,

a1=133,a2=-4,x=2y=-1z=3

Put all the values in distance formula

the distance between these two parallel lines =133-(-4)22+(-1)2+32

=13+12314

=25314

Hence, Option (C) is the correct answer.


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