Line passes through a point parallet to parametric equations

[megr_ftw]

Homework Statement

Consider the line which passes through the point P(-1, 4, 3), and which is parallel to the line x = 1 + 3t, y = 2 + 4t, z = 3 + 5t
Find the point of intersection of this new line with each of the coordinate planes:
xy-plane: ( , , 0 )
xz-plane: ( , 0 , )
yz-plane: ( 0 , , )

Homework Equations

The Attempt at a Solution

I can figure out the whole intersection thing, I just don't understand how I'm suppose to make it parallel to a line at the same time.

 

[Mark44]

I think you're approaching this in the wrong order. First, find the parameteric equations for the line through (-1, 4, 3), then figure out where this line intersects the coordinate planes.

 

[tiny-tim]

I can figure out the whole intersection thing, I just don't understand how I'm suppose to make it parallel to a line at the same time.

Hi megr_ftw!

I don't understand what you're saying here …

an intersection is a point, it can't be parallel to a line (or to anything).

You're asked what is the line through P parallel to the given line.

 

[megr_ftw]

so how do i go about finding that line that is parallel but passes through the certain point?

 

[Gregg]

So the line in question is parallel to this line:

$$\hat{r}=\left( \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right) + \left( \begin{array}{c} 3 \\ 4 \\ 5 \end{array} \right)t$$

Since it passes through the given point as described it must be that the new line is this?

$$\hat{s}=\left( \begin{array}{c} -1 \\ 4 \\ 3 \end{array} \right) + \left( \begin{array}{c} 3 \\ 4 \\ 5 \end{array} \right)t$$

Intersection of the planes will be found in the instance that:

$$3t - 1 = 0$$

$$4t + 4 = 0$$

$$5t + 3 = 0$$

For yz-,xz- and xy-planes.

To find the co-ordinates the values found for t need be plugged into the line s.