Determining perpendicular planes

frusic
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Determine whether the planes are perpindicular

(-2, 1, 4) . (x-1, y, z+3) = 0 (PLANE A)
(1, -2, 1) . (x+3, y-5, z) = 0 (PLANE B)



Here's what I have figured out so far:

Plane A passes through (1,0,-3) and is perpendicular to (-2,1,4)
Plane B passes through (-3, 5, 0) and is perpendicular to (1, -2, 1)

I know that if I had to determine if they were parallel, (1,0,-3) and (-2,1,4) would have to be along the lines of (-1, 2, 4) and (2, -4, -8).

I'm not sure if I'm on the right track and missing something right in front of me, or completely lost altogether.
 
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Isn't the definition of two planes being perpendicular that their normal vectors are perpendicular?
 
Yes I'm sure, but I'm not sure what the numbers of perpindicular normal vectors look like.

Like, the example of parallel vectors I gave above - I recognize those as being parallel, but I'm not sure how to tell if something is perpendicular unless the vectors are drawn out.
 
Aren't (-2, 1, 4) and (1, -2, 1) the normal vectors?
 
If they're perpendicular then the angle between then is 90 degrees. Then their dot product is |normal vector 1||normal vector 2|*cos90 = |normal vector 1||normal vector 2|*0 = 0. So they're perpendicular if their dot product is 0.
 
Thanks Dick and JG, it makes sense! I think I was making it harder than it actually was :)
 
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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