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Find the distance between the point P (7,5,4) and the plane determined by points A (3,-2,5), B (5,6,7) and C (–1, –4,6).

Find the distance between the point P (7,5,4) and the plane determined by points A (3,-2,5), B (5,6,7) and C (–1, –4,6).

Grade:Upto college level

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
A palne passing through (3,-2,5) is a(x-3)+b(y+2)+c(z-5)=0
It also passes through points (5,6,7) and (-1,-4,6) hence
a+4b+c=0 and -4a-2b+c=0
solving above equations we get
\frac{a}{6}=\frac{b}{5}=\frac{c}{14}
so eqn of plane is 6x+5y+14z=78
The distance from P(x,y,z) to the plane ax+by+cz=d is:

|ax+by+cz-d| / sqrt(a^2 + b^2 + c^2)

Therefore |ax+by+cz-d| = |6(7) + 5(5) + 14(4) -78| =45.

sqrt(a^2 + b^2 + c^2) = sqrt (36 +25 +196) = sqrt257.

The distance is 45/sqrt257
Thanks and Regards
Shaik Aasif
askIITians Faculty

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