Prove that two vectors whose direction cosines are given by relation $a l + b m + c n = 0$ and $f m n + g n l + h l m = 0$ are perpendicular, if $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0.$

Open in App

Solution

$a l + b m + c n = 0$ ………. $(1)$
$f m n + g n l + h l m = 0$ ………… $(2)$
$(1) \Rightarrow n = \frac{- a l - b m}{c}$ Using This in $(2)$
$f m \frac{(- a l - b m)}{c} + g l \frac{(- a l - b m)}{c} + h l m = 0$
$\Rightarrow a g l^{2} + b f m^{2} + (a f + b g - c h) l m = 0$
$\Rightarrow a g (l / m) + b f (m / l) + (a f + b g + c h) = 0$
$\Rightarrow a g (l / m)^{2} + (a f + b g + c h) (l / m) + b f = 0$
whose roots are $\frac{l_{1}}{m_{1}}, \frac{l_{2}}{m_{2}}$
$∴ \frac{l_{1} l_{2}}{m_{1} m_{2}} = \frac{b f}{a g}$ ………. $(3)$

Similarly $(1) \Rightarrow l = \frac{- b m - c n}{a}$ using this in $(2)$

We get $\frac{n_{1} n_{2}}{m_{1} m_{2}} = \frac{b h}{c g}$ ………. $(4)$

Now,

$l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$ ( $∵$ lines are perpendicular)
$\frac{l_{1} l_{2}}{m_{1} m_{2}} + 1 + \frac{n_{1} n_{2}}{m_{1} m_{2}} = 0$
$\Rightarrow \frac{b f}{a g} + 1 + \frac{b h}{c g} = 0$
Multiply equation by $g / b$
$\Rightarrow f / a + g / b + h / c = 0$ .

Suggest Corrections

Similar questions

The straight lines whose direction cosines are given by al + bm + cn = 0, fmn + gnl + hlm = 0 are perpendicular, if

a l + b m + c n = 0, f m n + g n l + h l m = 0

a l + b m + c n = 0

f m n + g l n + h l m = 0

f, g, h, a, b, c

l, m, n

f = g = h = 1

a l + b m + c n = 0

f m n + g l n + h l m = 0

f, g, h, a, b, c

l, m, n

f = g = h = 1

Join BYJU'S Learning Program