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Question

The shortest distance between the lines x-10=y+1-1=z1 and x+y+z+1=0, 2x-y+z+3=0 is:


A

1

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B

12

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C

13

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D

12

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Solution

The correct option is C

13


Explanation for the correct option:

Given: the lines x-10=y+1-1=z1 and x+y+z+1=0, 2x-y+z+3=0

Plane through the line of intersection of lines x+y+z+1=0 and 2x-y+z+3=0, is x+y+z+1+λ2x-y+z+3=0 (plane passing through the intersection of two lines)

x+y+z+1+2λx-λy+λz+3λ=0

x(1+2λ)+y(1-λ)+z(1+λ)+1+3λ=0 -------------1

It should be parallel to given line x-10=y+1-1=z1 -------------2

Therefore direction cosine and normal plane vector will be perpendicualr

0(1+2λ)+-1(1-λ)+1(1+λ)=0λ-1+λ+1=0λ=0 (a1a2+b1b2+c1c2=0)

And 1will become

x(1+2(0))+y(1-0)+z(1+0)+1+3(0)=0x+y+z+1=0

And line pass through (1,-1,0).

Shortest distance from a point (x1,y1,z1) to line ax+by+cz+d=0 is ax1+by1+cz1+da2+b2+c2

Shortest distance of (1,-1,0) from this plane =1-1+0+112+12+12

=13

Thus, the shortest distance is 13.

Hence, option C is correct.


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