Question

consider the following planes.
5x - 4y + z = 1, 4x + y - 5z = 5
(a) find parametric equations for the line of intersection of the planes. (use the parameter t.)
(x(t), y(t), z(t)) = (5 + 19t, - 4 + 29t,1 + 21t)×
(b) find the angle between the planes. (round your answer to one decimal place.)
748√

Explanation:

Step1: Find the direction vector of the line of intersection

The normal vectors of the planes $5x - 4y+z = 1$ and $4x + y-5z = 5$ are $\vec{n_1}=\langle5,-4,1
angle$ and $\vec{n_2}=\langle4,1,-5
angle$. The direction vector $\vec{v}$ of the line of intersection is the cross - product of the normal vectors: $\vec{v}=\vec{n_1}\times\vec{n_2}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\5&-4&1\\4&1&-5\end{vmatrix}$$

=\vec{i}(20 - 1)-\vec{j}(- 25-4)+\vec{k}(5 + 16)=\langle19,29,21
angle$.

Step2: Find a point on the line of intersection

Set $z = 0$. Then we have the system of equations

$$\begin{cases}5x-4y=1\\4x + y=5\end{cases}$$

. Multiply the second equation by $4$: $16x+4y = 20$. Add it to the first equation: $(5x-4y)+(16x + 4y)=1 + 20$, $21x=21$, so $x = 1$. Substitute $x = 1$ into $4x + y=5$, we get $4+y=5$, $y = 1$. So a point on the line is $(1,1,0)$.

Step3: Write the parametric equations

Using the point $(1,1,0)$ and direction vector $\langle19,29,21
angle$, the parametric equations are $x(t)=1 + 19t$, $y(t)=1+29t$, $z(t)=21t$.

Answer:

$(x(t),y(t),z(t))=(1 + 19t,1+29t,21t)$