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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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hey dennis u...(X+2)/3 –  sundar nataraj Сундар Oct 7 '12 at 15:05
    
please change like that –  sundar nataraj Сундар Oct 7 '12 at 15:05
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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative "Prove" to be rude when asking for help; please consider rewriting your post. –  Dennis Gulko Oct 7 '12 at 15:07
    
Thank you , I felt ur comment much useful..sure I willl DO! –  sundar nataraj Сундар Oct 7 '12 at 15:08

2 Answers 2

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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