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

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

#### 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.

Concept: Distance of a Point from a Plane
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