# Given 2 Vectors, how do I determine if they are parallel, skew or intersecting?

In general, how do I determine if 2 vectors (usually with variables like $\lambda$ and $\mu$). For example:

I did

$3+\lambda = 3 - 2\mu$

$\lambda = -2\mu$

Sub into $-2 + 3 \lambda = 2-6\mu$

$-2 + 3 (-2\mu) = 2-6\mu$

$-2 - 6 \mu = 2 - 6 \mu$

$-2 = 2$ What?

Anyway when doing such questions, how do I determine if 2 vectors with variables are parallel, skew or intersecting. If a lines are parallel, won't they definitely intersect? What is skew?

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You have $L_1$ is a line parametrized by ${\bf r}_1(t)=\langle -2,3,1 \rangle + \langle 3,1,-3 \rangle t$

and $L_2$ is parametrized by ${\bf r}_2(s)=\langle 2,3,4 \rangle + \langle -6,-2,6 \rangle s$

First, look at the direction vectors of both lines: ${\bf r}_1'(t) = \langle 3,1,-3 \rangle$ and ${\bf r}_2'(s) = \langle -6,-2,6 \rangle$. Since these vectors differ by a scalar multiple, they are parallel. This means both lines "go" in the same direction.

Thus the lines are either parallel or the same line (with different parametrizations).

Parallel lines do not intersect, whereas the same line will share all points. So setting the parametrizations equal to each other and solving equations will allow you to determine whether they intersect or not.

${\bf r}_1(t)={\bf r}_2(s)$ means that $-2+3t=2-6s$, $3+t=3-2s$, $1-3t=4+6s$. The second equation says $t=-2s$. Plugging this into the first equation says $-2+3(-2s)=2-6s$ so that $-4-6s=-6s$ so that $-4=0$. Since this is impossible, the equations have no solution. This means the lines are parallel.

If the direction vectors had not been parallel we would have had either intersecting or skew lines. Skew lines are non-parallel non-intersecting lines.

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Rewrite $L_2$ as: $2\bf{i}+3j+4k-2\mu(3i+j-3k)$. Then note that the direction vector of $L_1$ is a scalar multiple of that of $L_2$. Hence $L_1$ and $L_2$ are parallel.

What did shows that the lines do not intersect.

Skew lines are non-intersecting, non-parallel lines.

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