Question

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

The plane contains the point $(1,2,-5)$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 $Q=(1,2,-5)$ and the line determined by $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4t⟩$is $-4x+8y+6z+18=0$

Step-by-step solution

Given information

The plane that contains the point $Q=(1,2,-5)$and the line determined by $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4t⟩$

Calculation

The goal is to determine the plane equation that is determined by the given conditions.

The given point $Q=(1,2,-5)$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.

$$\begin{gathered} 3+7t=1 \\ -2+t=2 \\ 3+4t=-5 \end{gathered}$$

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:

$$\begin{gathered} \overrightarrow{Q_{0}Q}=⟨1-3,2-(-2),-5-3⟩ \\ =⟨-2,4,-8⟩ \end{gathered}$$

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:

$\mathbf{N}=\left|\begin{array}{c}ijk\\ 714\\ -24-8\end{array}\right|$

$$\begin{gathered} =⟨-24,48,30⟩\text{(Scale the cross product by}\frac{1}{6}\text{)} \\ =⟨-4,8,5⟩ \end{gathered}$$

Calculation

The equation of the plane is:

$$\begin{gathered} -4(x-1)+8(y-2)+6(z+5)=0 \\ -4x+8y+6z+4-16+30=0\text{(Simplify)} \\ -4x+8y+6z+18=0\text{(Combine like terms)} \end{gathered}$$

The equation of the plane that contains the point$Q=(1,2,-5)$and the line determined by $\mathbf{r}\left(t\right)=⟨3+7t,-2+t,3+4t⟩$is $-4x+8y+6z+18=0$