Question
Prove that two vectors whose direction cosines are given by relations al + bm + cn = 0
and $\mathrm{fmn}+\mathrm{gnl}+\mathrm{hlm}=0$ are perpendicular if $\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$
✓
Answer
Given, al + bm + cn = 0 …(1) and fmn + gnl + hlm = 0 …..(2)
From (1), $\mathrm{n}=-\left(\frac{a l+b m}{c}\right) \ldots . .(3)$
Substituting this value of n in equation (2), we get (fm + gl)∙[latex]-\left(\frac{a l+b m}{c}\right)[/latex] + hlm = 0
$\begin{aligned} & \left.\therefore-\left(a f l m+b f m^2+a g\right)^2+b g l m\right)+c h l m=0 \\ & \left.\therefore a g\right|^2+(a f+b g-c h) l m+b f m^2=0 \ldots(4)\end{aligned}$
Note that both l and m cannot be zero, because if l = m = 0, then from (3), we get
$n=0$, which is not possible as $1^2+m^2+n^2=1$.
Let us take m # 0.
Dividing equation (4) by $\mathrm{m}^2$, we get
$a g\left(\frac{l}{m}\right)^2+(a f+b g-c h)\left(\frac{l}{m}\right)+b f=0 \ldots(5)$
This is quadratic equation in $\left(\frac{l}{m}\right)$.
If $I_1, m_1, n_1$ and $I_2, m_2, n_2$ are the direction cosines of the two lines given by the equation
(1) and $(2)$, then $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$ are the roots of the equation (5).
From the quadratic equation (5), we get
product of roots $=\frac{l_1}{m_1} \cdot \frac{l_2}{m_2}=\frac{b f}{a g}$
$\therefore \frac{l_1 l_2}{m_1 m_2}=\frac{(f / a)}{(g / b)}$
$\therefore \frac{l_1 l_2}{(f / a)}=\frac{m_1 m_2}{(g / b)}$
Similarly, we can show that,
$\frac{l_1 l_2}{(f / a)}=\frac{n_1 n_2}{(h / c)}$
$\therefore \frac{l_1 l_2}{(f / a)}=\frac{m_1 m_2}{(g / b)}=\frac{n_1 n_2}{(h / c)}=\lambda \quad \ldots$ (Say)
$\therefore l_1 l_2=\lambda\left(\frac{f}{a}\right), m_1 m_2=\lambda\left(\frac{g}{b}\right), n_1 n_2=\lambda\left(\frac{h}{c}\right)$
Now, the lines are perpendicular if
$l_1 l_2+m_1 m_2+n_1 n_2=0$
i.e. if $\lambda\left(\frac{f}{a}\right)+\lambda\left(\frac{g}{b}\right)+\lambda\left(\frac{h}{c}\right)=0$
i.e. if $\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$.
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