Equation of Plane Perpendicular to Given Line l Through Point (-1, -4, 3)

[lennes]

5. Let l be the line x = −2 + 2t, y = 1 − 2t, z = −3 + t.
Find an equation of the plane W perpendicular to l through the point (−1, −4, 3).

i mean to give
2(x+1)-2(y+4)+(z-3)
=2x-2y+z-9 = 0

but the answer given is: 2x-2y+z-3 = 0

is the answer given wrong, or am I doing something wrong?

 

[dacruick]

I don't think either of those answers are right. The point (-1,-4,3) has to lie on the plane. That means that the point will have to satisfy the equation of the plane, and it seems not to do so.

 

[dacruick]

Also, I do know that the scalar equation of a plane (the equations you've given ax +by + cz + D= 0), has coefficients a, b, and c as the normal vector.

 

[lennes]

Sorry i think i made a mistake copying my answer

i mean to give
2(x+1)-2(y+4)+(z-3)
=2x-2y+z-9 = 0

 

[dacruick]

can you show your work for me?

 

[lennes]

Let l be the line x = −2 + 2t, y = 1 − 2t, z = −3 + t.
(a) Find an equation of the plane W perpendicular to l through the point (−1, −4, 3).

since the line perpendicular to W has direction of (2, -2, 1) i made it the normal to the plane.
n = (2, -2, 1)
so the point normal form i got
2(x - (-1)) -2(y - (-4))+1(z - (3)) = 0

2x+2 -2y-8 +z+3 = 0
2x-2y+z +2-11 = 0
2x-2y+z-9 =0

 

[HallsofIvy]

That looks perfectly good to me. Also note that 2(-1)- 2(-4)+ 3- 9= -2+ 8+ 3- 9= 0, so the given point is on this plane while 2(-1)- 2(-4)+ 3- 3= -2+ 8+ 3- 3= 6, not 0, so the given point is not on that plane.