Intersection of Planes?

Question

Find the equation of the line that is the intersection of the planes x=y=z=1 and y+z=0. (Please show work)

I made a typo...it should be x+y+z=1

Answer 1

Find the equation of the line that is the intersection of the planes x + y + z = 1 and y + z = 0.

x + y + z - 1 = 0
y + z = 0

The directional vector v, of the line of intersection of the two planes will be normal to the normal vectors n1 and n2, of the two planes. Take the cross product.

v = n1 X n2 = <1, 1, 1> X <0, 1, 1> = <0, -1, 1>

Now find a point on the line of intersection. It is immediately obvious that the point P(1, 0, 0) is on the line of intersection.

With the point P and the directional vector v we can write the equation of the line of intersection.

L(t) = P + tv = <1, 0, 0> + t<0, -1, 1>
where t is a scalar ranging over the real numbers

Answer 2

ok in the event that they're parallel they do no longer intersect in any respect, in the event that they do no longer seem to be parallel they intersect as a line. what different determination is there? 3 planes can intersect as a factor or as a line. replaced into the question approximately 3 planes? i would not confront your instructor yet could recheck the question and if it asks approximately 2 planes intersecting i could ask for a proof, for the reason which you don't get it. he could have made a mistake whilst writing out the attempt or have been given it out of a defective text fabric or you have gotten misinterpret.

Answer 3

if your asking the intersection between planes x=y=z=1 and y+z = 0, there is no intersection between these four planes namely x=1, y=1, z=1, and y+z=0.

the first three planes are clearly intersecting at point (1,1,1), thus by substitution,
y+z=0
1+1=0
2=0 is senseless.

but if it would be x+y+z=1 and y+z=0
we will have
y=-z. then by substitution we will have
x+(-z)+z=0
then the plane is x=0