Find the equation of plane whose points are (1,2,3) and (5,-2,4) is perpendicular to x+y-z=0.
Solution:
We know that for perpendicular
A(x-x1)+B(y-y1)+C(z-z1) = 0
Passing through (1,2,3):
A(x-1)+B(y-2)+C(z-3) = 0 ...(1)
Now again the plane (1) is passing through the point (5,-2,4):
A(5-1)+B(-2-2)+C(4-3) = 0
4A-4B+C = 0 ...(a)
And also perpendicular to x+y-z=0
Therefore
A+B-C=0 ...(b)
Now from a and b :
4A-4B+1C=0
1A+1B-1C=0
(A/4-1)-(B/-4-1)+(C/4+4)=K
(A/3)+(B/5)+(C/8)=K
So,
A = 3K , B = 5K and C = 8K
Putting values in eq(1) we get,
3K(x-1)+5K(y-2)+8K(z-3)=0
3x-3+5y-10+8z-24=0
3x+5y+8z-37=0
3x+5y+8z=37 ans...
Hence the equation of plane is 3x+5y+8z-37=0.
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