Question

Determine whether the given line is parallel to, intersects, or lies in the given plane. If the line is parallel to the plane, calculate its distance from the plane. If the line intersects the plane, find the point and angle at which they intersect.

$\mathrm{r}\left(t\right)=⟨t,-5-2t,-1+3t⟩$ and $4x-5y+2z=5$

Short answer

The angle between line $\mathrm{r}\left(t\right)=⟨t,-5-2t,-1+3t⟩$ and the plane $4x-5y+2z=5$ is $\theta =52.83°$

Step-by-step solution

Given information

The line $\mathrm{r}\left(t\right)=⟨t,-5-2t,-1+3t⟩$and the plane $4x-5y+2z=5$

Calculation

The line $\mathrm{r}\left(t\right)=⟨t,-5-2t,-1+3t⟩$and the plane $4x-5y+2z=5$

The goal is to figure out whether a given line is parallel to, intersects with, or lies within a particular plane. Calculate the dot product of the distance and normal vectors to determine the behavior of the line.

The direction vector of the line $\mathrm{r}\left(t\right)=⟨t,-5-2t,-1+3t⟩$is ${\mathbf{d}}_1=⟨1,-2,3⟩$and the normal vector of the plane $4x-5y+2z=5\$$is ${\mathbf{N}}_2=⟨4,-5,2⟩$

The dot product of direction vector and the normal vector is:

$$\begin{gathered} {\mathbf{d}}_1·{\mathbf{N}}_2=⟨1,-2,3⟩·⟨4,-5,2⟩ \\ =4+10+6 \\ =20 \end{gathered}$$

The dot product is greater than zero. As a result, the plane and the line cross.

The point of intersection is obtained for the value of parameter $t=t_0$ such that the line $\mathbf{r}\left(t_0\right)$ is on the plane.

The coordinates $\left(x\left(t_0\right),y\left(t_0\right),z\left(t_0\right)\right)$ are given by:

$$\begin{gathered} x\left(t_0\right)=t_0 \\ y\left(t_0\right)=-5-2t_0 \\ z\left(t_0\right)=-1+3t_0 \end{gathered}$$

Calculation

The points satisfy the plane $4x-5y+2z=5$

Therefore, on substitution the following equation is obtained.

$$\begin{gathered} 4t_0-5\left(-5-2t_0\right)+2\left(-1+3t_0\right)=5\text{(Substitution)} \\ 4t_0+25+10t_0-2+6t_0=5 \\ 20t_0+23=5 \\ 20t_0=-18(\text{Simplify)} \\ {t}_0=-\frac{18}{20} \\ =-\frac{9}{10} \end{gathered}$$

The point of intersection is:

$\left(x\left(t_0\right),y\left(t_0\right),z\left(t_0\right)\right)=\left(-\frac{9}{10},-\frac{32}{10},-\frac{37}{10}\right)$