Where does the line through A(1,0,1) and B(4,-2,2) intersect the plane x+y+z=6?

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Homework Statement


Where does the line through A(1,0,1) and B(4,-2,2) intersect the plane x+y+z=6?


Homework Equations


r= r_0 + tv


The Attempt at a Solution



The line through A and B is the vector AB and can be found by subtracting the A components from the B components. So B-A= <3,-2,1>.

They give x+y+z=6 so the components of that vector is <1,1,1>.

Not sure where to go from here...
 
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Put the components of your parametric equation for the line, e.g x=1+3t into the equation for the plane and solve for t, please? The normal vector to the plane doesn't have much to do with it.
 
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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