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# 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.

Question
Equations asked 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.

## Answers (1) 2020-11-15
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
$$\displaystyle\vec{{{n}_{{1}}}}={\left(\begin{array}{c} {1}\\{1}\\{1}\end{array}\right)}$$
The normal vector for the plane x + z = 0 is
$$\displaystyle\vec{{{n}_{{2}}}}={\left(\begin{array}{c} {1}\\{0}\\{1}\end{array}\right)}$$
Step 3
The cross product of the normal vectors is
$$\displaystyle\vec{{v}}={\left|{\vec{{{n}_{{1}}}}{X}\vec{{{n}_{{2}}}}}\right|}={\left|{\begin{array}{ccc} \vec{{i}}&\vec{{j}}&\vec{{k}}\\{1}&{1}&{1}\\{1}&{0}&{1}\end{array}}\right|}$$
$$\displaystyle=\vec{{i}}{\left({1}-{0}\right)}-\vec{{j}}{\left({1}-{1}\right)}+\vec{{k}}{\left({0}-{1}\right)}$$
$$\displaystyle=\vec{{i}}-\vec{{k}}$$
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 $$\displaystyle{r}_{{0}}={\left({0},{2},{0}\right)}$$.
That is, $$\displaystyle{r}_{{0}}={0}\vec{{i}}+{2}\vec{{j}}+{0}\vec{{k}}={2}\vec{{j}}$$
Step 4
The vector equation is given by,
$$\displaystyle{r}={r}_{{0}}+{t}{v}$$
$$\displaystyle{r}={2}\vec{{j}}+{t}{\left(\vec{{i}}-\vec{{k}}\right)}$$
$$\displaystyle{r}={t}\vec{{i}}+{2}\vec{{j}}-{t}\vec{{k}}$$
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

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