Question

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

The plane contains the point (−4, 1, 3) and is normal to the line determined by

Short answer

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

Step-by-step solution

Given information

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

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 $\mathbf{r}\left(t\right)=⟨-4+t,3+5t,2-3t⟩$as the three different solutions are obtained from the following equation.

$$\begin{gathered} -4+t=-4 \\ 3+5t=1 \\ 2-3t=3 \end{gathered}$$

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

The vector parallel to the plane of interest is:

$$\begin{gathered} \overrightarrow{Q_{0}Q}=⟨-4+4,1-3,3-2⟩ \\ =⟨0,-2,1⟩ \end{gathered}$$

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

Calculation

The normal vector is:

$\mathbf{N}=\left|\begin{array}{c}ijk\\ 15-3\\ 0-21\end{array}\right|$

$=⟨-1,2,-2⟩$

The equation of the plane is:

$$\begin{gathered} -1(x+4)+2(y-1)-2(z-3)=0 \\ -x+2y-2z-4-2+6=0\text{(Simplify)} \\ -x+2y-2z=0\text{(Combine like terms)} \end{gathered}$$

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