1.

Find the equation of the plane passing through the point `(2,-3,5)` and parallel to the points `3x-7y-2z=5`. Also, the find the distance between the two planes.

Answer» The given plane is `3x-7y-2z-5=0`……………..(i)
Let the required plane parallel to the given plane be
`3x-7y-2z+lambda=0` for some real number `lambda`…………(ii)
Since, the plane (ii) passes through the point `(2,-3,5)` , we have
`(3xx3)-7 xx (-3)-(2xx5)+lambda=0 rArr (6+21-10)+lambda=0`
`rArr lambda=-17`.
`therefore` the equation of the required plane is `(3x-7y-2z-17=0`...........(iii)
Now, the planes (i) and (ii) are parallel and the point (2,-3,5) lies on (iii).
`therefore` distance between the two planes.
= distance between the point (2,-3,5) and the plane (i)
`=(|(3xx2)-7xx(-3)-(2xx5)-5|)/(sqrt(3^(2)+(-7)^(2)+(-2)^(2))`
`=(|6+21-10-5|)/(sqrt(9+49+4))`
`=12/sqrt(62) xx sqrt(62)/sqrt(62) = 12/62sqrt(62)`
`=6/31sqrt(62)` units.
Hence, the distance between the two planes is `6/31 sqrt(62)` units.


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