Question

consider the following planes. x + y+z = 1, x + 5y + 5z = 1. (a) find parametric equations for the line of intersection of the planes. (use the parameter t.) (x(t), y(t), z(t)) = ( ) (b) find the angle between the planes. (round your answer to one decimal place.)

Explanation:

Step1: Find normal vectors

The normal vector of the plane $x + y+z = 1$ is $\vec{n_1}=\langle1,1,1
angle$, and the normal vector of the plane $x + 5y+5z = 1$ is $\vec{n_2}=\langle1,5,5
angle$.

Step2: Calculate the dot - product

The dot - product $\vec{n_1}\cdot\vec{n_2}=1\times1 + 1\times5+1\times5=1 + 5+5 = 11$.

Step3: Calculate the magnitudes

$|\vec{n_1}|=\sqrt{1^2 + 1^2+1^2}=\sqrt{3}$, $|\vec{n_2}|=\sqrt{1^2 + 5^2+5^2}=\sqrt{1 + 25+25}=\sqrt{51}$.

Step4: Find the angle between the planes

The angle $\theta$ between the two planes satisfies $\cos\theta=\frac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}|\times|\vec{n_2}|}=\frac{11}{\sqrt{3}\times\sqrt{51}}=\frac{11}{\sqrt{153}}\approx\frac{11}{12.369}\approx0.89$. Then $\theta=\arccos(0.89)\approx27.1^{\circ}$.

Step5: Find the line of intersection (parametric equations)

We solve the system of equations

$$\begin{cases}x + y+z = 1\\x + 5y+5z = 1\end{cases}$$

. Subtract the first equation from the second: $(x + 5y+5z)-(x + y+z)=1 - 1$, which gives $4y + 4z = 0$, so $y=-z$. Let $z=t$, then $y=-t$. Substitute $y=-t$ and $z = t$ into the first equation $x+(-t)+t = 1$, we get $x = 1$.
So the parametric equations of the line of intersection are $x(t)=1$, $y(t)=-t$, $z(t)=t$.

Answer:

(a) $(x(t),y(t),z(t))=(1,-t,t)$
(b) $27.1^{\circ}$