SOLUTION: Where does the line (x; y; z) = (4; 2; 1) + t(-2;-1; 0), (t is an element of the Real numbers), intersect the plane with general equation 2x - y + 3z = 6? any help would be ap

Algebra ->  Linear-equations -> SOLUTION: Where does the line (x; y; z) = (4; 2; 1) + t(-2;-1; 0), (t is an element of the Real numbers), intersect the plane with general equation 2x - y + 3z = 6? any help would be ap      Log On


   

Question 940706: Where does the line (x; y; z) = (4; 2; 1) + t(-2;-1; 0), (t is an element of the Real numbers), intersect the plane with general
equation 2x - y + 3z = 6?
any help would be appreciated cheers
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
So the line is then,
(4-2t,2-t,1%2B0t)=(4-2t,2-t,1)
Substitute this into the plane equation and solve for t.
2x-y%2B3z=6
2%284-2t%29-%282-t%29%2B3%281%29=6
8-4t-2%2Bt%2B3=6
9-3t=6
-3t=-3
t=1
So then,
(4-2,2-1,1)=(2,1,1) is the intersection point.