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#### Raerin

##### Member

- Oct 7, 2013

- 46

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.

- Thread starter Raerin
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- Thread starter
- #1

- Oct 7, 2013

- 46

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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- Jan 26, 2012

- 4,197

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- #3

- Oct 7, 2013

- 46

It's point 2 minus point 1:

(3, 0, 0) - (0, 0, -1) = (3, 0, 1)?

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- #4

- Jan 26, 2012

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Excellent! Now what does your candidate equation look like?It's point 2 minus point 1:

(3, 0, 0) - (0, 0, -1) = (3, 0, 1)?

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- #5

- Oct 7, 2013

- 46

(x, y, z) = (3, 0, 0) + t(3, 0, 1)Excellent! Now what does your candidate equation look like?

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- #6

- Jan 26, 2012

- 4,197

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