Question

Use Exercise 17 to find the angle between the indicated lines and planes in Exercises 46 and 47. $\mathrm{r}\left(t\right)=⟨3+5t,2-t,4-3t⟩$and $-10x+2y+6z=7$

Short answer

The angle between line $r\left(t\right)=⟨3+5t,2-t,4-3t⟩$ and the plane $-10x+2y+6z=7$ is $\theta =\frac{\pi }{2}$

Step-by-step solution

Given information

Consider the line $\mathrm{r}\left(t\right)=⟨3+5t,2-t,4-3t⟩$ and the plane $-10x+2y+6z=7$

Calculation

The goal is to determine the angle formed by the specified 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:

$\theta ={\cos }^{-1}\left(\frac{{\mathbf{d}}_1·{\mathbf{N}}_2}{||{\mathbf{d}}_1||||{\mathbf{N}}_2||}\right)$

The direction vector of the line $r\left(t\right)=⟨3+5t,2-t,4-3t⟩$ is ${\mathbf{d}}_1=⟨5,-1,-3⟩$ and the normal vector of the plane $-10x+2y+6z=7$ is ${\mathbf{N}}_2=⟨-10,2,6⟩$

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

$$\begin{gathered} {\mathbf{d}}_1·{\mathbf{N}}_2=⟨5,-1,-3⟩·⟨-10,2,6⟩ \\ =-50-2-18 \\ =-70 \end{gathered}$$

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

Calculation

The angle between the line and the plane is:

$$\begin{gathered} \theta ={\cos }^{-1}\left(\frac{⟨5,-1,-3⟩·⟨-10,2,6⟩}{|⟨5,-1,-3⟩|\left|\right|-10,2,6⟩}\right)\text{(Substitution)} \\ ={\cos }^{-1}\left(\frac{|-50-2-18|}{\sqrt{25+1+9}\sqrt{100+4+36}}\right)\text{(Dot Product)} \\ ={\cos }^{-1}\left(\frac{70}{\sqrt{35}\sqrt{140}}\right)\text{(Simplify)} \\ ={\cos }^{-1}\left(\frac{70}{70}\right) \\ ={\cos }^{-1}\left(1\right) \\ =0^0 \end{gathered}$$

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 $0°\mathrm{Is}90°$

The angle between line $r\left(t\right)=⟨3+5t,2-t,4-3t⟩$and the plane $-10x+2y+6z=7$is $\theta =\frac{\pi }{2}$