Question

x+y-z=3 and 2x+y+3z=4
find the vector and parametric equations of intersection

Answer 1

what is the normal vector of each plane?

Answer 2

idk

Answer 3

I won't prove it right now, but for a plane of equation\[ax+by+cz=d\]the normal vector is\[\vec n=\langle a,b,c\rangle\]

Answer 4

now considering that the normal vector is perpendicular to each plane, then the cross-product of each normal vector should be parallel to both planes, and hence the line of intersection of those plane
make sense?

Answer 5

i got the vector what about the other part

Answer 6

that part I am more rusty on....

Answer 7

we just need a point of intersection of the two planes, the question is how to find it

Answer 8

I found a point of intersection but my method is less than kosher...

Answer 9

solve the system\[x+y-z=3\]\[2x+y+3z=4\]subtract eqn 1 from eqn 2\[x+4z=1\implies x=1-4t|t\in\mathbb R\]let t=0, the above then imlpies\[t=0\implies x=1\implies z=0\]eqn 1 implies then that\[\implies y=2\]which is also a solution to equation 2, so the point \(P=(1,2,0)\) is in both planes
the formula for the line of intersection is\[\vec r(t)=P+t\vec n\]