Question

Find the equations of the planes determined by the given conditions.

The plane contains the point $(-4,1,3)$and the line determined

by$\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4t⟩$

Short answer

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

Step-by-step solution

Given information

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

Calculation

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 $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4t⟩$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 $\mathbf{d}=⟨7,1,4⟩$and the point $Q_0=(3,-2,3)$is on the line $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4t⟩$

The vector parallel to the plane of interest is:

$$\begin{gathered} \overline{Q_{0}Q}=⟨-4-3,1-(-2),3-3⟩ \\ =⟨-7,3,0⟩ \end{gathered}$$

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

The normal vector is:

$\mathbf{N}=\left|\begin{array}{c}ijk\\ 714\\ -730\end{array}\right|$$$\begin{gathered} =⟨-12,-28,28⟩ \\ \left.=⟨-3,-7,7⟩\text{(Scale the cross-product by}\frac{1}{4}\right) \end{gathered}$$

Calculation

The equation of the plane is:

$$\begin{gathered} -3(x+4)-7(y-1)+7(z-3)=0 \\ -3x-7y+7z-12+7-21=0\text{(Simplify)} \\ -3x-7y+7z-26=0\text{(Combine like terms)} \end{gathered}$$

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