Thread: Intersection of line with YZ plane

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)

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?

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.
$\displaystyle
\left|\begin{array}{rcl}0& =& 1 + m\\ y &=& 3\\ z&=& 2 + 2m\end{array}\right.$

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.