# [SOLVED]How to find the intersection point between two lines

#### Raerin

##### Member
How to find the intersection point between two lines?

line 1: r = (3,1,-1) + s(1,2,3)
line 2: r = (2,5,0) + s(1,-1,1)

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
line 1: r = (3,1,-1) + s(1,2,3)
line 2: r = (2,5,0) + s(1,-1,1)
Hey Raerin!

Let's use a different parameter in both of those line equations.

line 1: r = (3,1,-1) + s(1,2,3)
line 2: r = (2,5,0) + t(1,-1,1)

We're looking for an $s$ and a $t$, such that the resulting r is the same...

Your post looks a bit out of context now.

#### Petrus

##### Well-known member
Hey Raerin!

Let's use a different parameter in both of those line equations.

line 1: r = (3,1,-1) + s(1,2,3)
line 2: r = (2,5,0) + t(1,-1,1)

We're looking for an $s$ and a $t$, such that the resulting r is the same...

Your post looks a bit out of context now.
Haha! It Was a typo I see!! I Was trying and trying, i Was like how can $$\displaystyle 3+s= 2+s$$

Regards,
$$\displaystyle |\pi\rangle$$

#### Raerin

##### Member
Hey Raerin!

Let's use a different parameter in both of those line equations.

line 1: r = (3,1,-1) + s(1,2,3)
line 2: r = (2,5,0) + t(1,-1,1)

We're looking for an $s$ and a $t$, such that the resulting r is the same...

Your post looks a bit out of context now.
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So it's not possible for both parameters to be s? Then there's a mistake in the question. I guess I don't need help anymore. Thanks!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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So it's not possible for both parameters to be s? Then there's a mistake in the question. I guess I don't need help anymore. Thanks!
It's not really a mistake in the question.
The general form of a line equation is $\vec r = \vec a + s \vec d$.
However, when you intersect 2 different lines with such an equation, you have to realize that the parameters $s$ in those 2 line equations are distinct.