Solve 3D Vector Equation: x-int=3, z-int=-1; (x,y,z)=(p1,p2,p3)+t(d1,d2,d3)

[Raerin]

Question:
Write a vector equation which has an x-intercept of 3 and a z-intercept of -1

I want it in the form of (x,y,z) = (p1, p2, p3) + t(d1, d2, d3)

I don't have a clear idea on how to solve it.

 

[Ackbach]

So, if you're finding the equation of a line, and you're given two points on the line, that should determine the line. The $(d_{1},d_{2},d_{3})$ vector you've given is going to be the direction vector. Any idea how you could get that?

 

[Raerin]

So, if you're finding the equation of a line, and you're given two points on the line, that should determine the line. The $(d_{1},d_{2},d_{3})$ vector you've given is going to be the direction vector. Any idea how you could get that?

It's point 2 minus point 1:
(3, 0, 0) - (0, 0, -1) = (3, 0, 1)?

 

[Ackbach]

It's point 2 minus point 1:
(3, 0, 0) - (0, 0, -1) = (3, 0, 1)?

Excellent! Now what does your candidate equation look like?

 

[Raerin]

Excellent! Now what does your candidate equation look like?

(x, y, z) = (3, 0, 0) + t(3, 0, 1)

 

[Ackbach]

(x, y, z) = (3, 0, 0) + t(3, 0, 1)

Right. And check it out: $t=0$ yields your $x$ intercept, and $t=-1$ yields your $z$ intercept. So, this is the straight line through those two points.