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Show that the points (1, –1, 3) and (3, 4, 3) are equidistant from the plane 5x + 2y – 7z + 8 = 0

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#### Solution

Let p1 and p2 be the distances of points `hati-hatj+3hatk and 3hati+4hatj+3hatk` from `bar r.(5hati+2hatj-7hatk)+8=0`

The distance of the point A with position vector a from the plane `barr.barn` = p is given by

`d=|bara.barn-p|/|barn|`

`therefore p_1=|(hati-hatj+3hatk).(5hati+2hatj-7hatk)-(-8)|/sqrt(5^2+2^2+(-7)^2)`

=`|1(5)-1(2)+3(-7)+8|/sqrt(25+4+49)`

=`|5-2-21+8|/sqrt(78)=|-10|/sqrt78=10/sqrt78`

`and p_2=|()()-(-8)|/sqrt(5^2+2^2+(-7)^2)`

=`|3(5)+4(2)+3(-7)+8|/sqrt(25+4+49)`

=`|15+8-21+8|/sqrt78`

=`10/sqrt78`

∴ p_{1} = p_{2}

Hence, points are equidistant from the plane.

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