Finding a point where a line in parametric form meets a plane

jcfor3ver
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Homework Statement



Here is the problem:

x=y-1=2z

and the equation of the plane is 4x-y+3z=8





Homework Equations





The Attempt at a Solution



Ya so i got the normal line to be <1,1,-1/2> but i do not know where to go from here? help please?
 
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hi jcfor3ver! :wink:

you want the point that lies on both lines?

then just solve the pair of simultaneous equations! :smile:​
 

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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