Chapter 11.4, Problem 52E

a.

To determine

To find: The angle between given two planes.

Answer to Problem 52E

  $\theta =0^0$ .

Explanation of Solution

Given information:

Equations of two plane is given as,

  $\begin{array}{l}2x+4y-2z=1\\ -3x-6y+3z=10\end{array}$

Concepts used:

An equation of plane given in the general form of $ax++by+cz+d=0$ has a normal vector to the plane as $\text{n}=\langle a,b,c\rangle$ .

Angle of intersection between two planes having normal vectors ${\text{n}}_1{\text{ and n}}_2$ is given by-

  $\cos \theta =\frac{\left|{\text{n}}_1\cdot {\text{n}}_2\right|}{\Vert {\text{n}}_1\Vert \Vert {\text{n}}_2\Vert }$

Calculation:

Equations of two plane is given as,

  $\begin{array}{l}2x+4y-2z=1\\ -3x-6y+3z=10\end{array}$

So, normal vectors to these plane will be given as −

  ${\text{n}}_1=\langle 2,4,-2\rangle$

  ${\text{n}}_2=\langle -3,-6,3\rangle$

Now angle between these two plane will be given by −

  $\begin{array}{l}\cos \theta =\frac{\left|{\text{n}}_1\cdot {\text{n}}_2\right|}{\Vert {\text{n}}_1\Vert \Vert {\text{n}}_2\Vert }\\ \Rightarrow \cos \theta =\frac{\left|\langle 2,4,-2\rangle \cdot \langle -3,-6,3\rangle \right|}{\sqrt{2^2+4^2+{\left(-2\right)}^2}\sqrt{{\left(-3\right)}^2+{\left(-6\right)}^2+3^2}}\\ \Rightarrow \cos \theta =\frac{\left|-6-24-6\right|}{\sqrt{24}\sqrt{54}}\\ \Rightarrow \cos \theta =\frac{36}{\sqrt{24}\sqrt{54}}\\ \Rightarrow \theta =0^0\end{array}$

Hence, angle between given two planes is $\theta =0^0$ .

b.

To determine

To find: The parametric equation of line of intersection of given two plane.

Answer to Problem 52E

Planes do not intersect.

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

From part $a$ , angle between line of intersection of two plane is $0^0$ . This means these planes either never intersects or are same plane.