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

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