Question 1048763: Find the general form of the equation of the plane with the given characteristics.
The plane passes through the points, (-4,-7,7) and (-2,-1,-6)
and is perpendicular to the plane x-3y+5z=-1.
*Please explain in detail!!!*
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! A normal vector to the plane x-3y+5z=-1 is < 1,-3, 5 >.
Since the points (-4,-7,7) and (-2,-1,-6) are on the desired plane, a vector parallel to the desired plane is < -2--4, -1--7, -6-7 > = < 2,6, -13 >.
Let < a,b,c > be the normal vector of the desired plane.
< 2,6, -13 >*< a,b,c > = 0, or
2a + 6b - 13c = 0 <------Equation A
Also, < 1,-3,5 >*< a,b,c > = 0, or
a - 3b + 5c = 0, <-------Equation B
as per the given that the two planes are perpendicular.
Multiplying Equation B by 2 and adding with Equation A gives
 .
Multiplying Equation B by -2 and adding with Equation A gives
 .
===> The normal vector < a,b,c > = c < 3/4, 23/12,1 >.
Choosing c = 12, let the normal be < 9, 23, 12 >.
Therefore, the desired plane has the equation
9(x+2)+23(y+1)+12(z+6) = 0.
I leave the simplifying to you.
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