# 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.

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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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