91Ó°ÊÓ

The line $r\left(t\right)=⟨5+2t,6-t,4+5tâŸ\copyright$ and the plane $3y+5z=-4$

The goal is to determine the angle formed by the indicated line and the plane. The acute angle between the line's direction vector and the plane's normal vector is the counterpart of the angle between the lines intersecting the plane.

The angle formed by intersecting lines on the plane is given by:

$\mathrm{θ}={\cos }^{-1}\left(\frac{{\mathbf{d}}_1·{\mathbf{N}}_2}{||{\mathbf{d}}_1||âˆ\pounds {\mathbf{N}}_2‖}\right)$

The direction vector of the line $r\left(t\right)=⟨5+2t,6-t,4+5tâŸ\copyright$is $d_1=⟨2,-1,5âŸ\copyright$and the normal vector of the plane $3y+5z=-4$is ${\mathbf{N}}_2=⟨0,3,5âŸ\copyright$

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

$$\begin{gathered} {\mathbf{d}}_1·{\mathbf{N}}_2=⟨2,-1,5âŸ\copyright ·⟨0,3,5) \\ =0-3+25 \\ =22 \end{gathered}$$

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

The angle between the line and the plane is:

$$\begin{gathered} \mathrm{θ}={\cos }^{-1}\left(\frac{⟨2,-1,5)·⟨0,3,5âŸ\copyright }{‖⟨2,-1,5âŸ\copyright ‖‖0,3,5âŸ\copyright âˆ\pounds }\right)\text{(Substitution)} \\ ={\cos }^{-1}\left(\frac{0-3+25}{\sqrt{4+1+25}\sqrt{0+9+25}}\right)\text{(Dot Product)} \\ ={\cos }^{-1}\left(\frac{22}{\sqrt{30}\sqrt{34}}\right)(\text{Simplify)} \\ ={\cos }^{-1}\left(\frac{11}{\sqrt{255}}\right) \\ ≈46.46 \end{gathered}$$

The acute angle between the line's direction vector and the normal vector to the plane is the counterpart of the angle between the line and the plane.

The complementary angle of $46.46°$is $43.54°$

The angle between line $\mathrm{r}\left(t\right)=⟨5+2t,6-t,4+5t)$ and the plane $3y+5z=-4$ is $\mathrm{θ}=43.54°$