The plane equations are x + y + z = 2 and x + z = 0.
(a) The parametric equations for the line of intersection of the planes
x + y + z = 2 and x + z = 0.
(b)The symmetric equations.
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
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 .
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
(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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