Phoebe

2020-11-14

a) Find parametric equations for the line. (Use the parameter t.)
The line of intersection of the planes
x + y + z = 2 and x + z = 0
(x(t), y(t), z(t)) =
b) Find the symmetric equations.

Raheem Donnelly

Step 1
Given:
The plane equations are x + y + z = 2 and x + z = 0.
To find:
(a) The parametric equations for the line of intersection of the planes
x + y + z = 2 and x + z = 0.
(b)The symmetric equations.
Step 2
(a)
Consider the planes x + y + z = 2 and x + z = 0.
To find the vector equation of the line of intersection, we need to find the cross product v of the normal vectors of the given planes and a point on the line of intersection.
The normal vector for the plane x + y + z = 2 is
$\stackrel{\to }{{n}_{1}}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$
The normal vector for the plane x + z = 0 is
$\stackrel{\to }{{n}_{2}}=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)$
Step 3
The cross product of the normal vectors is
$\stackrel{\to }{v}=|\stackrel{\to }{{n}_{1}}X\stackrel{\to }{{n}_{2}}|=|\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ 1& 1& 1\\ 1& 0& 1\end{array}|$
$=\stackrel{\to }{i}\left(1-0\right)-\stackrel{\to }{j}\left(1-1\right)+\stackrel{\to }{k}\left(0-1\right)$
$=\stackrel{\to }{i}-\stackrel{\to }{k}$
To find a point on the line of intersection, put z = 0 in both the plane equations,
x + y + z = 2 and x + z = 0, we get
x + y = 2 and x = 0
That is x = 0 and y = 2.
Therefore, the point of intersection is ${r}_{0}=\left(0,2,0\right)$.
That is, ${r}_{0}=0\stackrel{\to }{i}+2\stackrel{\to }{j}+0\stackrel{\to }{k}=2\stackrel{\to }{j}$
Step 4
The vector equation is given by,
$r={r}_{0}+tv$
$r=2\stackrel{\to }{j}+t\left(\stackrel{\to }{i}-\stackrel{\to }{k}\right)$
$r=t\stackrel{\to }{i}+2\stackrel{\to }{j}-t\stackrel{\to }{k}$
Therefore, the parametric equations for the line of intersection of the planes are
x = t , y = 2, z = -t
Step 5
(b) The symmetric equations:
To find the symmetric equation, we solve each of the parametric equations for t and then set them equal.
t = x, y = 2, t = -z
Setting them equal gives us the symmetric form:
x = -z and y=2

xleb123

(a)$\begin{array}{cc}\hfill x\left(t\right)& =-t,\hfill \\ \hfill y\left(t\right)& =2,\hfill \\ \hfill z\left(t\right)& =t.\hfill \end{array}$
(b)$\begin{array}{cc}\hfill x+t& =0,\hfill \\ \hfill y& =2,\hfill \\ \hfill z-t& =0.\hfill \end{array}$
Explanation:
a) To find the parametric equations for the line of intersection of the planes $x+y+z=2$ and $x+z=0$, we can express the variables $x$, $y$, and $z$ in terms of the parameter $t$.
From the equation $x+z=0$, we have $x=-z$.
Substituting this expression for $x$ in the equation $x+y+z=2$, we get $-z+y+z=2$. Simplifying, we obtain $y=2$.
Therefore, the parametric equations for the line are:
$\begin{array}{cc}\hfill x\left(t\right)& =-t,\hfill \\ \hfill y\left(t\right)& =2,\hfill \\ \hfill z\left(t\right)& =t.\hfill \end{array}$
b) To find the symmetric equations for the line, we can eliminate the parameter $t$ from the parametric equations.
From the equation $x\left(t\right)=-t$, we have $t=-x$. Substituting this expression for $t$ in the equation $z\left(t\right)=t$, we get $z=-x$. Thus, we have $z=-x$.
The symmetric equations for the line of intersection are:
$\begin{array}{cc}\hfill x+t& =0,\hfill \\ \hfill y& =2,\hfill \\ \hfill z-t& =0.\hfill \end{array}$

Jazz Frenia

a) Start by solving the following system of equations to determine the parametric equations for the line connecting the planes $x+z=0$ and $x+y+z=2$.
$\begin{array}{c}\hfill x+y+z=2\\ \hfill x+z=0\end{array}$
From the second equation, we can isolate $x$ in terms of $z$ as $x=-z$. Substituting this expression for $x$ in the first equation, we get:
$-z+y+z=2⟹y=2$
Thus, we have $x=-z$ and $y=2$, which gives us the parametric equations:
$\begin{array}{c}\hfill x\left(t\right)=-t\\ \hfill y\left(t\right)=2\\ \hfill z\left(t\right)=t\end{array}$
where $t$ is the parameter.
b) To find the symmetric equations for the line, we can express the line using the vector form. The direction vector of the line is given by the coefficients of $t$ in the parametric equations, which in this case is $⟨-1,0,1⟩$.
Using a point on the line, such as $\left(0,2,0\right)$, we can write the vector equation as:
$⟨x,y,z⟩=⟨0,2,0⟩+t⟨-1,0,1⟩$
Expanding this equation gives us the symmetric equations:
$\frac{x-0}{-1}=\frac{y-2}{0}=\frac{z-0}{1}$
Simplifying further, we can write:
$\frac{x}{-1}=\frac{z}{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=2$
Thus, the symmetric equations for the line of intersection are:
$\begin{array}{c}\hfill x=-z\\ \hfill y=2\end{array}$

Do you have a similar question?