Easy Vector Question | Picture Included | Homework Help

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Homework Statement



Check the picture.

Homework Equations





The Attempt at a Solution



I'm sure it's very simple but I just can't seem to get it. I've been sitting here staring at the page for a while now with no joy. Check the picture for my attempt.
Thanks very much for any help, greatly appreciated.
 

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Can anyone help with this?
 
well you are correct in saying that \hat{b}=\frac{\vec{ab}}{|ab|} and that \hat{c}=\frac{\vec{ac}}{|ac|}.

and if \vec{ad} = t( \hat{b} - \hat{c}) what does that tell you about the direction of that vector?

How would you find the vector bc?

then how do you get the angle between two vector lines?
 
Ah I see it now so \vec{ad} = t ( \hat{cb} )
Therefore \vec{ad} is perpendicular to \hat{cb}
Therefore \vec{ad} is the perpendicular bisector of the the line bc.
Thank you very much for your help!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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