Vector lines, ratio of vector lines

thoradicus
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http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_w09_qp_31.pdf

Homework Statement


6i


Homework Equations


r=a+λb


The Attempt at a Solution


first i find AB which is (2,2,-2)
the equation if line is then (1,0,-1)+λ(2,2,-2)
i made it so that λ is 0.5, which then OM is (2,1,-2), as its the mid point.

i found AC to be (3,-3,3)

the equation of AC is (1,0,-1)+λ(3,-3,3)
which then i made lambda to be 1/3...but is this the right working? i did that and i can't get the awnser.
 
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N is 2/3 of the way from A to C so is closer to C than A. Take \lambda= 2/3, not 1/3.
 
gah, didnt notice that. I am such an idiot >< thanks alot.
 
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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