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Question
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`
∴ p1 = p2
Hence, points are equidistant from the plane.
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