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If the distance between the plane Ax+2y+z=d and the plane containing the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5 is `sqrt6`, then |d| is equal to

A

3

B

4

C

6

D

1

Text Solution

Verified by Experts

The correct Answer is:
C
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Find the shortest distance between the following pair of lines : (x-1)/(2) = (y-2)/(3)=(z-3)/(4),(x-2)/(3)=(y-3)/(4)=(z-5)/(5)

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HIMALAYA PUBLICATION-THREE DIMENSIONAL GEOMETRY-QUESTION BANK
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  4. A line from the origin meets the lines (x-2)/1 = (y-1)/(-2) = (z+1)/1 ...

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  10. If the straight line, x =1 + s, y = 3-lambdas, z = 1+ lambdas and x=t/...

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  12. If the angle theta between the (x+1)/1=(y-1)/2=(z-2)/2 and the plane 2...

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  14. Let L be the line of intersection of the planes: 2x + 3y +z = 1 and ...

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  15. If the line passing through the point (5,1,a) and (3,b,1) crosses the ...

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  16. If the straight lines: (x-1)/k=(y-2)/2=(z-3)/3 and (x-2)/3 =(y-3)/k ...

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  17. Let the line (x-2)/3=(y-1)/(-5)=(z+2)/2 lie in the plane x+3y-alpha z+...

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  18. A line AB in three-dimensional space makes angles 45^@ and 120^@ with ...

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  19. The shortest distance between the lines (x-1)/2 = (y-2)/3 = (z-3)/4 an...

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  20. The distance between the line vec r = 2i+2j+3k + lambda (i-j+4k) and t...

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