# Calculus

Find the scalar equation of a plane that is perpendicular to plane r=(-2,1,3)+s(0,1,1)+t(-1,0,3) and also intersects it along the line (x-3)/2=(y+1)/-3,z=1

1. 0
2. 0
3. 22
1. A normal to r is
(0,1,1) cross (-1,0 3)
or
(3, -1, 1) , (I assume you know how to find a cross-product)

So the plane must look like,
3x - y + z = c , where c is a constant

from the line (x-3)/2 = (y+1)/-3 , z=1
the point (3,-1 ,1) must lie on the line, so it must also lie in the plane

3(3) - (-1) + 1 = c
c = 11

3x - y + z = 11

1. 0
2. 0
posted by Reiny

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