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Question

If the distance between the plane ax4y+2z=k and the plane containing the lines x23=y34=z45 and x34=y45=z56 is, 26 then |k|3 is equal to

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Solution

Distance between two planes = distance of the point (2, 3, 4) from the plane 2x4y+2z=k.

Let l, m, n be d.c.'s of normal to the plane of given lines.

l.3+k.4+n.5=0

l.4+m.5+n.6=0

l2425=m1820=n1516

l1=m2=n1

But the plane ax4y+z=k must be parallel to this plane.

a1=42=11a=2

distance between the planes =26

= distance of the planes 2x4y+2zk=0 from the point (2, 3, 4)

=412+8k4+16+4=|k|26

4×6=|k|

|k|3=243=8

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