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[SOLVED] 3D vector equation

Raerin

Member
Oct 7, 2013
46
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

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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

Member
Oct 7, 2013
46
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

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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

Member
Oct 7, 2013
46

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
(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.