# Thread: Lines Planes and vectors in R3 Problem

1. ## Lines Planes and vectors in R3 Problem

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...

2. Are we to assume that "L" is the line parallel to the given vector?

If so, you are misinterpreting the problem. The point (-2, 3, 1) cannot be on the line. If it were, there would be an infinite number of planes passing through that line (and point). L must be any other line, parallel to that vector, not containing that point.

Write $\displaystyle x= x_0- 2t$, $\displaystyle y= y_0+ t$, $\displaystyle z= z_0+ 3t$ where you are free to choose $\displaystyle (x_0, y_0, z_0)$ as long as it gives a line not passing through (-2, 3, 1). For example, taking $\displaystyle x_0= y_0= z_0= 0$ gives $\displaystyle x= -2t$, $\displaystyle y= t$, $\displaystyle z= 3t$ which does NOT pass through (-2, 3, 1) because if $\displaystyle y= t= 3$ then $\displaystyle z= 3t= 9\ne 1$.

You can find another vector in the plane by taking a vector from any given point on that line to (-2, 3, 1). For example, taking t= 0 gives (0, 0, 0) so another vector in the plane is <-2, 3, 1>. The cross product of that and the given vector is normal to the plane.