Thread: Intersection of line with YZ plane

Originally Posted by gva0324

Here I have a supposedly simple vector question but my answer seems to be wrong. What am I doing wrong?

A(1,3,2) u= i + 3j +k v= 3j-k

a) If vector AB = u - v what are coordinates of B

AB = OB - OA

OB = (u -v) + OA

OB = (2,3,4) ........... OK

b)i) Write down the parametric equation of the line L, passing thorugh the point A in the direction of AB.

L = OA + m(AB)

L = (1,3,2) + m(1,0,2) ........... <<<<<<<<<

ii) What is the value of m at which L cross the YZ plane.

L crosses the YZ plane when x= 0

(0,y,z) = (1,3,2) + m(1,0,2)

0 = 1 + m
y = 3
z= 2 + 2m

My problem here is that the z turns out to be zero. This is not possible since it says it crosses the YZ plane. What is wrong with this?

All your considerations are OK.


From the first equation you get m = -1. Then the point of intersection is P(0, 3, 0). The point of intersection is placed on the y-axis. But points of the y-axis belong to the yz-plane too.