Thread: Vector equation of a line and points of intersection

Dear forum members, I don't understand why I am getting the wrong answer

A line goes through the points (1,-3,1) and (13,-12,4). Does it cross the x-axis? If so, at what point?

My solution:

I call the points A and P, respectively. I form the direction vector AP, which becomes ( -4 -5 -1) (sorry for the nonstandard vector representation, please pretend this is a column).

The equation of the line is then

r=(1,-3,1) +t(-4 -5 -1)

I know that at the point where the line crosses the x-axis, the y and z -coordinates are zero.

hence

x= 1-4t
y=-3-5t
z=1-t

y and z must equal zero, so I solve for t to see whether it will be the same for both. I get

for y t=-3/5
for z t=1

Hence, I would conclude that the line does not cross the x-axis, but the answer in my book says it does. Please help me find, where I went wrong.

Thank you. All help is much appreciated.

Originally Posted by Coach

Dear forum members, I don't understand why I am getting the wrong answer

A line goes through the points (1,-3,1) and (13,-12,4). Does it cross the x-axis? If so, at what point?

My solution:

I call the points A and P, respectively. I form the direction vector AP, which becomes ( -4 -5 -1) (sorry for the nonstandard vector representation, please pretend this is a column).

The equation of the line is then

r=(1,-3,1) +t(-4 -5 -1)

I know that at the point where the line crosses the x-axis, the y and z -coordinates are zero.

hence

x= 1-4t
y=-3-5t
z=1-t

y and z must equal zero, so I solve for t to see whether it will be the same for both. I get

for y t=-3/5
for z t=1

Hence, I would conclude that the line does not cross the x-axis, but the answer in my book says it does. Please help me find, where I went wrong.

Thank you. All help is much appreciated.

the DR's of line AP should be 12,-9,3
hence the eqn. of line is r=(i-3j+k) +t(12i-9j+3k)
x= 1-12t
y=-3-9t
z=1+3t
on comparing y and z to zero we get t=-1/3
so it crosses the x axis

Your mistake is