91Ó°ÊÓ

The plane that contains the point $Q=(-4,1,3)$ and the line determined by $\mathbf{r}\left(t\right)=⟨-4+t,3+5t,2-3tâŸ\copyright$

The goal is to discover the plane equation that is determined by the given circumstances.

The given point $Q=(-4,1,3)$is not on the line $\mathbf{r}\left(t\right)=⟨-4+t,3+5t,2-3tâŸ\copyright$as the three different solutions are obtained from the following equation.

The direction vector of the line is $\mathbf{d}=⟨1,5,-3âŸ\copyright$and the point $Q_0=(-4,3,2)$is on the line $\mathbf{r}\left(t\right)=⟨-4+t,3+5t,2-3tâŸ\copyright$

The vector parallel to the plane of interest is:

The normal vector to the plane is the cross product of $\mathbf{d}=⟨1,5,-3âŸ\copyright$and the vector $\stackrel{→}{Q_{0}Q}$

The normal vector is:

$=⟨-1,2,-2âŸ\copyright$

The equation of the plane is:

The equation of the plane that contains the point $Q=(-4,1,3)$ and the line determined by $\mathbf{r}\left(t\right)=⟨-4+t,3+5t,2-3tâŸ\copyright$is $-x+2y-2z=0$