91Ó°ÊÓ

The plane that contains the point $Q=(-4,1,3)$ and the line determined by $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4tâŸ\copyright$

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 $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4tâŸ\copyright$as the three different solutions are obtained from the following equation.

$$\begin{gathered} 3+7t=-4 \\ -2+t=1 \\ 3+4t=3 \end{gathered}$$

The direction vector of the line is $\mathbf{d}=⟨7,1,4âŸ\copyright$and the point $Q_0=(3,-2,3)$is on the line $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4tâŸ\copyright$

The vector parallel to the plane of interest is:

$$\begin{gathered} \stackrel{¯}{Q_{0}Q}=⟨-4-3,1-(-2),3-3âŸ\copyright \\ =⟨-7,3,0âŸ\copyright \end{gathered}$$

The normal vector to the plane is the cross product of $\mathbf{d}=⟨7,1,4âŸ\copyright$and the vector $\stackrel{→}{Q_{0}Q}$

The normal vector is:

The equation of the plane is:

The equation of the plane that contains the point $Q=(-4,1,3)$and the line determined by $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4tâŸ\copyright$is $-3x-7y+7z-26=0$