# How that the Points (1, 1, 1) and (-3, 0, 1) Are Equidistant from the Plane r(3i+4j-12k)+13=0 - Mathematics and Statistics

Show that the points (1, 1, 1) and (-3, 0, 1) are equidistant from the plane bar r (3bari+4barj-12bark)+13=0

#### Solution

Let p1 and p2 be the distances of (1, 1, 1) and (-3, 0, 1) from the plane  barr

(3hati+4hatj-12hatk)+13=0

cosider p_1=|((hati+hatj+hatk).(3hati+4hatj-12hatk)+13)/sqrt(3^2+4^2+(-12)^2)|

=|(1(3)+1(4)+1(-12)+13)/sqrt(9+16+144)|=|(3+4-12+13)/13=8/13.....(i)

also p_2=|((13hati+0hatj+hatk).(3hati+4hatj-12hatk)+13)/sqrt(3^2+4^2+(-12)^2)|

=|(-3(3)+0+1(-12)+13)/sqrt(9+16+144)|=|(-9-12+13)/13=8/13 ....(ii)

From (i) and (ii),

p1 = p2

Concept: Distance of a Point from a Plane
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