Chapter 10.4, Problem 47E

(a)

To determine

To find: The angle between the two planes $x+y-2z=0$ and $-4x-y+3z=0$ .

Answer to Problem 47E

The angle between the two planes $x+y-2z=0$ and $-4x-y+3z=0$ is $151.73°$ .

Explanation of Solution

Given information:

The equation of two planes $x+y-2z=0$ and $-4x-y+3z=0$ .

Calculation:

Compare the equation of the plane with $ax+by+cz=d$ .

So, the normal vector for plane $x+y-2z=0$ is $\langle 1,1,-2\rangle$ and the normal vector for plane $-4x-y+3z=0$ is $\langle -4,-1,3\rangle$ .

The angle between the two planes is equal to the angles between their normal vectors.

Use the formula for angle between two vectors $u$ and $v$ .

  $\cos \theta =\frac{u\cdot v}{\Vert u\Vert \Vert v\Vert }$

The dot product of both the vectors $\langle 1,1,-2\rangle$ and $\langle -4,-1,3\rangle$ .

  $\begin{array}{c}\langle 1,1,-2\rangle \cdot \langle -4,-1,3\rangle =1\times -4+1\times -1+\left(-2\right)\times 3\\ =-4-1-6\\ =-11\end{array}$

Use the formula for magnitude of vector $\langle 1,1,-2\rangle$ .

  $\begin{array}{c}\Vert \langle 1,1,-2\rangle \Vert =\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{1+1+4}\\ =\sqrt{6}\end{array}$

Use the formula for magnitude of vector $\langle -4,-1,3\rangle$ .

  $\begin{array}{c}\Vert \langle -4,-1,3\rangle \Vert =\sqrt{{\left(-4\right)}^2+{\left(-1\right)}^2+{\left(3\right)}^2}\\ =\sqrt{16+1+9}\\ =\sqrt{26}\end{array}$

Substitute all the values in the formula for angle.

  $\begin{array}{c}\cos \theta =\frac{\langle 1,1,-2\rangle \cdot \langle -4,-1,3\rangle }{\Vert \langle 1,1,-2\rangle \Vert \Vert \langle -4,-1,3\rangle \Vert }\\ =\frac{-11}{\sqrt{6}\times \sqrt{26}}\\ =\frac{-11\sqrt{39}}{78}\end{array}$

The solution for the trigonometric equation $\cos \theta =\frac{-11\sqrt{39}}{78}$ is $\theta =151.73°$ .

Therefore, the angle between the two planes $x+y-2z=0$ and $-4x-y+3z=0$ is $151.73°$ .

(b)

To determine

To find: The parametric equations of the line of intersection for the planes $x+y-2z=0$ and $-4x-y+3z=0$ .

Answer to Problem 47E

The parametric equations of the line of intersection of the planes $x+y-2z=0$ and $-4x-y+3z=0$ is $x=t$ , $y=5t$ and $z=3t$ .

Explanation of Solution

Given information:

The equation of two planes $x+y-2z=0$ and $-4x-y+3z=0$ .

Calculation:

Multiply the first equation of plane by $4$ and add to the second plane.

  $\begin{array}{c}4\left(x+y-2z\right)+\left(-4x-y+3z\right)=0\\ 4x+4y-8z-4x-y+3z=0\\ 3y-5z=0\\ 3y=5z\\ y=\frac{5}{3}z\end{array}$

Substitute $y=\frac{5}{3}z$ in the equation of plane $x+y-2z=0$ .

  $\begin{array}{c}x+y-2z=0\\ x+\frac{5}{3}z-2z=0\\ x=2z-\frac{5}{3}z\\ x=\frac{1}{3}z\end{array}$

Finally assume that $t=\frac{1}{3}z$ . This gives $z=3t$ .

Substitute $z=3t$ in the equations $x=\frac{1}{3}z$ and $y=\frac{5}{3}z$ .

  $\begin{array}{c}x=\frac{1}{3}z\\ =\frac{1}{3}\left(3t\right)\\ =t\end{array}$

and,

  $\begin{array}{c}y=\frac{5}{3}z\\ =\frac{5}{3}\left(3t\right)\\ =5t\end{array}$

Therefore, the parametric equations of the line of intersection of the planes $x+y-2z=0$ and $-4x-y+3z=0$ is $x=t$ , $y=5t$ and $z=3t$ .