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Line passing through $(1,0,5)$ and is perpendicular to the line $\frac{x+2}{3} = \frac{y-2}{4} = \frac{z+3}{5}$

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Hint: The sufficient and necessary conditions that two lines: $$L_1 :\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$ $$L_2: \frac{x-x_2}{a'}=\frac{y-y_2}{b'}=\frac{z-z_2}{c'}$$ are perpendicular is that $aa'+bb'+cc'=0$ so you can find $a',b',c'$ just by guessing. For example if $a'=2,b'=1,c'=-2$ we have $2*3+1*4+(-2)*5=0$ so the parametric equation of one prependicular line passing through $(1,0,5)$ would be $$\frac{x-1}{2}=\frac{y }{1}=\frac{z-5}{-2}$$

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I think, there are countless answer. The given line has a direction vector $u=(3,4,5)$. Put $v=(a,b,c)$ be a direction vector of the needing line. We have $$3a + 4b+5c=0.$$ We see that, $(-2, -1, 2$, $(-3, 1, 1)$, $(-5, 5, -1)$, $(-2, 4, -2)$, $(-1, 2, -1)$, ... are roots of this equation. That's mean, the equation $$3a + 4b+5c=0$$ has countless solutions. Therefore there are countless answers.

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Please explain. – Nate Eldredge Oct 7 '12 at 17:21

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