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.

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

Skilled2020-11-15Added 75 answers

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

$\overrightarrow{{n}_{1}}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$

The normal vector for the plane x + z = 0 is

$\overrightarrow{{n}_{2}}=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)$

Step 3

The cross product of the normal vectors is

$\overrightarrow{v}=\left|\overrightarrow{{n}_{1}}X\overrightarrow{{n}_{2}}\right|=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& 1& 1\\ 1& 0& 1\end{array}\right|$

$=\overrightarrow{i}(1-0)-\overrightarrow{j}(1-1)+\overrightarrow{k}(0-1)$

$=\overrightarrow{i}-\overrightarrow{k}$

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${r}_{0}=(0,2,0)$ .

That is,$r}_{0}=0\overrightarrow{i}+2\overrightarrow{j}+0\overrightarrow{k}=2\overrightarrow{j$

Step 4

The vector equation is given by,

$r={r}_{0}+tv$

$r=2\overrightarrow{j}+t(\overrightarrow{i}-\overrightarrow{k})$

$r=t\overrightarrow{i}+2\overrightarrow{j}-t\overrightarrow{k}$

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

The normal vector for the plane x + z = 0 is

Step 3

The cross product of the normal vectors is

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

That is,

Step 4

The vector equation is given by,

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

Skilled2023-05-23Added 181 answers

Jazz Frenia

Skilled2023-05-23Added 106 answers

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