Lines Planes and Vectors in 3 space

[gregory.weld]

If the line is parallel to the vector v = < -2,1,3>, what is the equation of the plane containing L and the point B = (-2,3,1)

A) - x + y + z = 6
B) 3x - 2y - z = 4
C) x + 6y -11z = 5
D) x + 5y - z = 12
E) 2x + 10y - 19z = 7

I know we can get the equation of the line because we have a point on the line and a vector parallel

So x = xo + at, y = yo + bt , z = zo + ct

x = -2 + -2t
y = 3 + t
z = 1 + 3t

I know the equation of a plane is a(x - xo) + b(y - yo) + c(z - zo) = 0. But the vector here <a,b,c> has to be normal to the plane...
Can someone help figure out how i find the answer

 

[Mark44]

If the line is parallel to the vector v = < -2,1,3>, what is the equation of the plane containing L and the point B = (-2,3,1)

A) - x + y + z = 6
B) 3x - 2y - z = 4
C) x + 6y -11z = 5
D) x + 5y - z = 12
E) 2x + 10y - 19z = 7

I know we can get the equation of the line because we have a point on the line and a vector parallel

So x = xo + at, y = yo + bt , z = zo + ct

x = -2 + -2t
y = 3 + t
z = 1 + 3t

I know the equation of a plane is a(x - xo) + b(y - yo) + c(z - zo) = 0. But the vector here <a,b,c> has to be normal to the plane...
Can someone help figure out how i find the answer

This is a tricky question. If all you know is a point on a plane and a vector that is parallel to a line that lies in the plane, there is not enough information to specify a unique plane.

However, since they give you the equations of 5 planes, then what you need to do is pick the one (if any) that works.

For each of the 4 planes that contains the given point (one plane doesn't), calculate the normal vector, and dot it with the direction vector of the line. If the result is 0, this means that the direction vector is perpendicular to the normal vector, so must lie in the plane. If the dot product isn't zero, the given line can't lie in the plane.

That's what I would do. I haven't worked it all the way through, so I can't say whether any of the given planes works.