HELP <20 mintues to Turn in online

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Tom McCurdy
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1.) Find vector parametric equation for the line through the point P=(1,4,–5) perpendicular to the plane –5x+3y–4z=–4.

Any help would be great
 
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Tom, you've been here long enough to know the rules!
 
I know I just need a hint
like how to find the normal vector or something
i wasn't hoping for the answer
 
The normal vector to a plane given by an equation ax+by+cz=0 is easy: all vectors (x,y,z) in the plane satisfy that equation, so are clearly normal to the vector... Replacing the 0 by a constant just shifts the plane, so the normal vector doesn't change.
 
thx guys for the hint but I'm to late... just have to get it done earlier next time
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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