Question

Suppose the direction cosines of
two lines are given by
al + bm + cn = 0 and
fmn + gln + hlm = 0 where
5 ,g, h,a,b,c are arbitrary constants
and l, m, n are direction cosines of
the lines. On the both of the above
information answer the following​

Answer 1

Co-ordinates (3D)

Complete question.

Show that if the straight lines whose direction cosines are given by $al+bm+cn=0$, $fmn+gnl+hlm=0$ be parallel, then one of the relations $\sqrt{af}\pm\sqrt{bg}\pm\sqrt{ch}=0$ is true.

Prove further that if the lines be at right angles, then

$\quad\quad\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$.

Proof.

Given relations are

$\quad\quad al+bm+cn=0$

$\quad\quad fmn+gnl+hlm=0$

Eliminating $l$ from the relations above, we get

$\quad fmn+gn\left(-\frac{bm+cn}{a}\right)+h\left(-\frac{bm+cn}{a}\right)m=0$

$\Rightarrow fmn-\frac{bgmn+cgn^{2}}{a}-\frac{bhm^{2}+chmn}{a}=0$

$\Rightarrow (af-bg-ch)mn-cgn^{2}-bhm^{2}=0$

$\Rightarrow bhm^{2}-(af-bg-ch)mn+cgn^{2}=0$

$\Rightarrow bh\left(\frac{m}{n}\right)^{2}-(af-bg-ch)\frac{m}{n}+cg=0\quad...(1)$

If $(l_{1},\:m_{1},\:n_{1})$ and $(l_{2},\:m_{2},\:n_{2})$ be the direction cosines of the given lines, then we can take $\frac{m_{1}}{n_{1}}$ and $\frac{m_{2}}{n_{2}}$ being the roots of equation $(1)$.

If the lines be parallel to each other, their direction cosines will be the same and the roots of the equation $(1)$ has to be equal.

For equal roots, we write

$\quad D=0$

$\Rightarrow \{-(af-bg-ch)\}^{2}-4bcgh=0$

$\Rightarrow (af-bg-ch)^{2}=4bcgh$

$\Rightarrow af-bg-ch=\pm 2\sqrt{bcgh}$

$\Rightarrow af=bg\pm 2\sqrt{bcgh}+ch$

$\Rightarrow af=(\sqrt{bg}\pm \sqrt{ch})^{2}$

$\Rightarrow \pm\sqrt{af}=\pm\sqrt{bg}+\pm\sqrt{ch}$

$\Rightarrow \boxed{\color{red}{\sqrt{af}\pm\sqrt{bg}\pm\sqrt{ch}=0}}$

Equation $(1)$ also gives

$\quad \frac{m_{1}m_{2}}{n_{1}n_{2}}=\frac{cg}{bh}$

$\Rightarrow \frac{m_{1}m_{2}}{\frac{g}{b}}=\frac{n_{1}n_{2}}{\frac{h}{c}}\quad.....(2)$

Similarly eliminating $n$ from the given relations, we get

$\quad \frac{l_{1}l_{2}}{\frac{f}{a}}=\frac{n_{1}n_{2}}{\frac{h}{c}}\quad.....(3)$

Combining $(2)$ and $(3)$, we have

$\quad \frac{l_{1}l_{2}}{\frac{f}{a}}=\frac{m_{1}m_{2}}{\frac{g}{b}}=\frac{n_{1}n_{2}}{\frac{h}{c}}=k\:(say)$

If the given two lines be at right angles, then

$\quad l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0$

$\Rightarrow \frac{kf}{a}+\frac{kg}{b}+\frac{kh}{c}=0$

$\Rightarrow \boxed{\color{red}{\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0}}$

This completes the proof.

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The plane passing through the origin and containing the lines whose direction cosines are proportional to $(1,\:-2,\:2)$ and $(2,\:3,\:1)$ passes through the point.

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