Finding the Cartesian Equation of a Perpendicular Line

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Homework Statement


Find the vector equation of the line that passes through the point Q(2,0,-5) and
is perpendicular to both the vectors m=(0,1,4) and n=(-2,-1,3).

Homework Equations


vector equation of a line: (x, y, z)=(x0,y0,z0) + t(a,b,c)
cartesian equation of a line: (x-x0)/a=(y-y0)/b=(z-z0)/c

The Attempt at a Solution


(x0,y0,z0)=(2,0,5)
To find (a,b,c)I know I can get two equations because the dot product of (a,b,c) with the two perpendicular lines equals zero:
b+4c=0
-2a-b+3c=0
But two equations isn't enough to solve for three variables. Also, shouldn't the point (2,0,-5) also dot product with u or v to equal zero, since it's on the same line?
Is it correct to assume that the points (0,1,4) and (-2,-1,3) are also points on the line? In which case I can easily find the direction of the line by subtracting one from the other.
 
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The cross product of <0, 1, 4> and <-2, -1, 3> will give you a vector that is perpendicular to both. That's the vector you need to write your line in either its vector form or in Cartesian form.
 
Man, I can't believe I didn't think of that myself. thanks for your help!
 
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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