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Show that if the lines with the directional cosines $(l, m, 0)$ and $(p, 0, q)$ are perpendicular then either $m = \frac {1}{\sqrt{p^2 + q^2}}$ or $q = \frac {1}{\sqrt {l^2 + m^2}}$.

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I am assuming you mean $m = \frac {1}{\sqrt {p^2 + q^2}}$ or $q = \frac {1}{\sqrt {l^2 + m^2}}$. – funktor Apr 10 '12 at 19:00
@BiditAcharya Yes you are right. Sorry for the typo. – Faisal Apr 10 '12 at 19:04
up vote 3 down vote accepted

Note here that when you write $(l,m,0)$ and $(p,0,q)$, these are the unit vectors along the respective lines.(Check on the definition of a direction cosine)
So, this'd mean that $$\sqrt{l^2+m^2}=1 ~~ \implies \frac1{\sqrt{l^2+m^2}}=1 \tag{1}$$ Similar reasoning allows us to deduce $$\sqrt{p^2+q^2}=1 ~~ \implies \frac1{\sqrt{p^2+q^2}}=1\tag{2} $$ Now note that $(l,m,0)$ and $(p,0,q)$ are perpendicular, so $$(l,m,0) \cdot(p,0,q)=0~~ \implies lp=0$$ One of them has to be zero right? Let's assume that $l=0$. Then from our result $(1)$, $m=1$ (it makes sense too, right? if $m \neq 1$ then our vector wouldn't be a unit vector would it?).

So one of our unit vectors reduces just to $(0,1,0)$, which is just a vector along the y-axis, or $\hat{\jmath}$.And what other unit vector do you know with the y-component $0$ (in the form $(p,0,q)$) and perpendicular to $\hat{\jmath}$? Obviously, it is the vector $\hat{k}$, or the vector $(0,0,1)$.

Now you should notice that from our deductions, $m=q=1$.
If you revisit $(1)$ and $(2)$ you have $$q=\frac1{\sqrt{l^2+m^2}}$$ and, $$m=\frac1{\sqrt{p^2+q^2}}$$

Hope it Helps!

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