# Thread: Is this Vector proof correct?

1. ## Is this Vector proof correct?

Prove that the point A $\displaystyle (i+2j-3k)$ lies on the line L1 $\displaystyle r = \left(\begin{array}{c}4\\-4\\3\end{array}\right) + \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

So what i have done is:

$\displaystyle \left(\begin{array}{c}1\\2\\-3\end{array}\right) = \left(\begin{array}{c}4\\-4\\3\end{array}\right) + \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

$\displaystyle \left(\begin{array}{c}1\\2\\-3\end{array}\right) - \left(\begin{array}{c}4\\-4\\3\end{array}\right) = \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

$\displaystyle \left(\begin{array}{c}-3\\6\\-6\end{array}\right) = \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

Therefore: $\displaystyle \mu = -3$

so Point A deos lie on the line L1.

am not sure if this is the correct way to prove it.

2. Hello Fedex,

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3. Yep. Spot on. The point is that the equations:

$\displaystyle \left(\begin{array}{c}-3\\6\\-6\end{array}\right) = \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

[/quote]

Give you the SAME value of mu. So the equations are consistent, so the point is on the line.

That's as good a way as any to prove it.

4. Hey Moo,

Thanks :-) I tried to copy and paste a formula, but I accidentally copied and pasted a picture. Pretty silly of me, but I sorted it now.

Thanks, though :-)

5. Thank You

The next part fo the question goes on to say:

The point C $\displaystyle (2i-k)$ lies on the line L2

Find the shortest distance from C to L1.

L1: $\displaystyle r = \left(\begin{array}{c}1\\2\\-3\end{array}\right) + \lambda \left(\begin{array}{c}4\\-5\\-3\end{array}\right)$

L2: $\displaystyle r = \left(\begin{array}{c}4\\-4\\3\end{array}\right) + \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

How would i go abou finding the shortest distance?