Asked by Kaur
find the perpendicular distance of point (4,2,-1) from the plane through the point (3,7,-2) (1,-1, 2) and (-2, 1, 3)
Answers
Answered by
mathhelper
label your points A(3,7,-2), B(1,-1, 2) and C(-2, 1, 3)
vector BA = (2,8,-4) or reduced to (1, 4, -2)
vector CA = (5, 6, -5)
a normal to these is ( 8, 5, 14) , (I assume you know how to find the cross-product )
The equation of the plane is 8x + 5y + 14z = c
(-2,1,3) lies on it, so
-16 + 5 + 42 = c = 31
Plane equation: 8x + 5y + 14z = 31
using the distance from a point to a plane formula
distance = |8(4) + 5(2) + 14(-1)|/√(64+25+196)
= 28/√285 units
or 28√285/285 units
check my arithmetic
vector BA = (2,8,-4) or reduced to (1, 4, -2)
vector CA = (5, 6, -5)
a normal to these is ( 8, 5, 14) , (I assume you know how to find the cross-product )
The equation of the plane is 8x + 5y + 14z = c
(-2,1,3) lies on it, so
-16 + 5 + 42 = c = 31
Plane equation: 8x + 5y + 14z = 31
using the distance from a point to a plane formula
distance = |8(4) + 5(2) + 14(-1)|/√(64+25+196)
= 28/√285 units
or 28√285/285 units
check my arithmetic
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