Chapter 11.4, Problem 49E

a.

To determine

To find: The angle between given two planes.

Answer to Problem 49E

  $\theta ={60.67}^0$ .

Explanation of Solution

Given information:

Equations of two plane is given as,

  $\begin{array}{l}3x-4y+5z=6\\ x+y-z=2\end{array}$

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,

  $\begin{array}{l}3x-4y+5z=6\\ x+y-z=2\end{array}$

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

  ${\text{n}}_1=\langle 3,-4,5\rangle$

  ${\text{n}}_2=\langle 1,1,-1\rangle$

Now angle between these two plane will be given by −

  $\begin{array}{l}\cos \theta =\frac{\left|{\text{n}}_1\cdot {\text{n}}_2\right|}{\Vert {\text{n}}_1\Vert \Vert {\text{n}}_2\Vert }\\ \Rightarrow \cos \theta =\frac{\left|\langle 3,-4,5\rangle \cdot \langle 1,1,-1\rangle \right|}{\sqrt{3^2+{\left(-4\right)}^2+5^2}\sqrt{1^2+1^2+{\left(-1\right)}^2}}\\ \Rightarrow \cos \theta =\frac{\left|3-4-5\right|}{\sqrt{50}\sqrt{3}}\\ \Rightarrow \cos \theta =\frac{6}{5\sqrt{6}}\\ \Rightarrow \theta ={60.67}^0\end{array}$

Hence, angle between given two planes is $\theta ={60.67}^0$ .

b.

To determine

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

Answer to Problem 49E

  $x=-t,y=16+8t,z=14+7t$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

Equations of two plane is given as,

  $\begin{array}{l}3x-4y+5z=6\mathrm{...}\text{eq(i)}\\ x+y-z=2\mathrm{...}\text{eq(ii)}\end{array}$

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

  ${\text{n}}_1=\langle 3,-4,5\rangle$

  ${\text{n}}_2=\langle 1,1,-1\rangle$

Since, line of intersection of two plane is parallel to the cross product of their normal vector.

  $\begin{array}{l}{\text{n}}_1\times {\text{n}}_2=\left|\begin{array}{ccc}i& j& k\\ 3& -4& 5\\ 1& 1& -1\end{array}\right|\\ =i\left(4-5\right)-j\left(-3-5\right)+k\left(3+4\right)\\ =-i+8j+7k\\ \text{So, the direction number for the line of intersection will be-}\end{array}$

  $a=-1,b=8,c=7$

Line of intersection of these two plane can be found by solving these two equation.

  $\begin{array}{l}\text{Performing eq(i)}+5\times \text{eq(ii),}\\ \frac{\begin{array}{l}3x-4y+5z=6\\ 5x+5y-5z=10\end{array}}{\begin{array}{l}8x+y+0=16\\ \Rightarrow y=16-8x\end{array}}\\ \text{Substitute }y=16-8x\text{ in eq(ii),}\\ x+16-8x-z=2\\ \Rightarrow z=14-7x\\ \text{when }x=0,y=16,z=14\end{array}$

Therefore, point $\left(0,16,14\right)$ lies on both plane means lies on line of intersection of two plane.

Now set of parametric equation of a line passing through $\left(0,16,14\right)$ and having direction numbers as $a=-1,b=8,c=7$ will be given as −

  $\begin{array}{l}x=x_1+at,y=y_1+bt,z=z_1+ct\\ \Rightarrow x=0-t,y=16+8t,z=14+7t\\ \Rightarrow x=-t,y=16+8t,z=14+7t\end{array}$

Hence, desired parametric equation of line of intersection is $x=-t,y=16+8t,z=14+7t$ .