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Find the equation of the plane, Need Help!?
P(1,1,3) Q(1,-3,5) R(3,0,2)
I keep getting wrong answers that only work for two points.
my current answer is 22x - 4y - 8z = -6
i used PQ=<0,4,-2>
PR<2,-1,-5>
UxV= <22,-4,-8>
and i used point P to find the equation
1 Answer
- Anonymous9 years agoFavourite answer
Your answer is wrong... You need to use the cross product of these vectors in order to find the slopes of the equation!
PQ = <0, -4, 2>
PR = <2, -1, -1>
Use the cross-product for these rays...
PQ x PR = <-4(-1) - (2)(-1), (2)(2) - (0)(-1), (0)(-1) - (-4)(2)>
= <6, 4, 8>
Hence, by taking point P and the above slope, you get...
6(x - 1) + 4(y - 1) + (z - 3) = 0
I hope this helps!
Source(s): Σ