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

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To solve for a vector equation with an x-intercept of 3 and a z-intercept of -1, the equation can be expressed as (x, y, z) = (3, 0, 0) + t(3, 0, 1). The direction vector is derived from the difference between the two intercept points, calculated as (3, 0, 0) - (0, 0, -1) = (3, 0, 1). This formulation confirms that when t=0, the x-intercept is reached, and when t=-1, the z-intercept is achieved. The resulting equation represents a straight line through the specified intercepts. Thus, the vector equation effectively describes the desired line in three-dimensional space.
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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.
 
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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?
 
Ackbach said:
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)?
 
Raerin said:
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?
 
Ackbach said:
Excellent! Now what does your candidate equation look like?

(x, y, z) = (3, 0, 0) + t(3, 0, 1)
 
Raerin said:
(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.
 
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