# Find the equation of the plane that contains the line r (t) = < 1 - t, 2 + 3 t, -3 + t > and the...

## Question:

Find the equation of the plane that contains the line {eq}\displaystyle r (t) = \langle 1 - t,\ 2 + 3 t,\ -3 + t \rangle {/eq} and the point {eq}\displaystyle (1,\ 2,\ 3) {/eq}.

## Equation of a Plane:

There are many different ways to determine a plane. We recall that a plane in space can be uniquely defined by a point {eq}P(x_0,y_0,z_0) {/eq} on the plane and a vector, {eq}\mathbf{n}=<a,b,c> {/eq}, that is perpendicular to the plane. This is known as a normal vector.

An equation for the plane through {eq}P(x_0,y_0,z_0) {/eq} perpendicular to {eq}\mathbf{n}=<a,b,c> {/eq} is

{eq}a(x-x_0)+b(y-y_0)+c(z-z_0)=0 {/eq}

To determine a plane we need a point on the plane and a vector normal to the plane. If the plane contains the point {eq}P(1,2,3) {/eq}, then we just need a normal vector.

Suppose that the plane also contains the line {eq}\mathbf r(t)=<1-t,2+3t,-3+t> {/eq}. We then have {eq}\mathbf r(0)=<1,2,-3> {/eq} giving us a second point of {eq}Q(1,2,-3) {/eq} on the plane. The direction of the line is {eq}\mathbf v=<-1,3,1> {/eq} is parallel to the plane and so is the vector {eq}\vec{PQ}=<1-1,2-2,-3-3>=<0,0,-6> {/eq}. It follows that a normal vector for the plane is

{eq}\begin{align} \mathbf n&=\mathbf v\times \vec{PQ}\\ &=<-1,3,1>\times <0,0,-6>\\ &=<3(-6)-0,0-(-1)(-6),0-0>\\ &=<-18,-6,0> \end{align} {/eq}

Using this normal vector and {eq}P {/eq} we get an equation of the plane as

{eq}-18(x-1)-6(y-2)+0(z-3)=0\\ 18(x-1)+6(y-2)=0 {/eq}