# Lines in 3D - Problem 3

We’re talking about lines in space. Let’s take a look at anther problem; where does the line <x, y, z> equals <6, 4, 10> plus t times <1, 2, -5> intersect the x,y plane? I’ve drawn a graph here. A quick sketch that shows the <x, y, z> space, the line, the point that it passes through <6, 4, 10>. And I wanted to point out that the x,y plane is down here.

Another way of describing the x,y plane is to call it z equals 0. Of course, the reason for that is that every point on the xy plane has a z coordinate of 0. So that’s an identifying characteristic of the plane. What I’m going to do is I’m going to return this into parametric equations. So I’ll get x equals 6 plus t, y equals 4 plus 2t and z equals 10 plus -5t. I need to find out for what parameter value t, z equals 0, because that’s where the line’s going to cross, the x,y plane.

So I set this equal to 0. I get 10 equals 5t, t equals 2. So I need to plug t equals 2 into these two equations. I get x equals 6 plus 2, 8, and y equals 4 plus, 2 times 2, 4, also 8. And so my coordinates are, <8, 8, 0>. It was really that easy. <8, 8, 0> is the point where this line crosses the x, y plane.

If you wanted to find where it crosses other planes, just use the equation of that plane. Like for example the y, z plane would be equals 0, or the x,z plane would be y equals 0. Just substitute 0 for the appropriate coordinate. Find the parameter value and then find the other coordinates, like we did here.

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