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

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Discussion Overview

The discussion revolves around formulating a vector equation for a line in three-dimensional space, specifically one that has an x-intercept of 3 and a z-intercept of -1. Participants explore how to express this equation in the form (x,y,z) = (p1,p2,p3) + t(d1,d2,d3).

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks for help in writing a vector equation with specified intercepts.
  • Another participant suggests that two points on the line can determine the line and questions how to derive the direction vector from these points.
  • A participant calculates the direction vector as the difference between the two points: (3, 0, 0) and (0, 0, -1), resulting in (3, 0, 1).
  • Following this, a participant presents a candidate equation for the line based on the derived points and direction vector.
  • It is noted that specific values of the parameter t yield the desired intercepts, confirming the line's correctness through the specified points.

Areas of Agreement / Disagreement

Participants appear to agree on the method of deriving the vector equation and the correctness of the proposed equation, though no explicit consensus on the final form is stated.

Contextual Notes

The discussion does not address potential limitations or assumptions regarding the derivation of the direction vector or the specific conditions under which the equation holds.

Raerin
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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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