# 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,

Then take the url of the pic and use the [img] things

Otherwise, you can click on "manage attachments", right below options, which are below the window where you type your message.

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?