Prove that the two vectors whose direction cosines are given by relations al + bm + cn = 0 and fmn + gnl + hlm = 0 are perpendicular, if `"f"/"a" + "g"/"b" + "h"/"c" = 0`

#### Solution

Given: al + bm + cn = 0 ...(1)

and fmn + gnl + hlm = 0 ...(2)

From (1), n = - `(("al + bm")/"c")` ...(3)

Substituting this value of n in equation (2), we get

(fm + gl).`[- ("al + bm")/"c"] + "hlm" = 0`

∴ - (aflm + bfm^{2} + agl^{2} + bglm) + chlm = 0

∴ agl^{2} + (af + bg - ch)lm + bfm^{2} = 0 ...(4)

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 l^{2} + m^{2} + n^{2} = 1

Let us take m ≠ 0.

Dividing equation (4) by m^{2}, we get

`"ag"(1/"m"^2) + ("af" + "bg" - "ch")(1/"m") + "bf" = 0` ....(5)

This is quadratic equation in `(1/"m")`

If l_{1}, m_{1}, n_{1} and l_{2}, m_{2}, n_{2} are the direction cosines of the two lines given by the equation (1) and (2), then `"l"_1/"m"_1` and `"l"_2/"m"_2` are the roots of the equation (5).

From the quadratic equation (5), we get

product of roots = `"l"_1/"m"_1 . "l"_2/"m"_2 = "bf"/"ag"`

∴ `("l"_1"l"_2)/("m"_1"m"_2) = ("f"//"a")/("g"//"b")`

∴ `("l"_1"l"_2)/("f"//"a") = ("m"_1"m"_2)/("g"//"b")`

Similarly, we can show that,

`("l"_1"l"_2)/("f"//"a") = ("n"_1"n"_2)/("h"//"c")`

∴ `("l"_1"l"_2)/("f"//"a") = ("m"_1"m"_2)/("g"//"b") = ("n"_1"n"_2)/("h"//"c") = lambda` ...(Say)

∴ `"l"_1"l"_2 = lambda ("f"/"a"), "m"_1"m"_2 = lambda ("g"/"b"), "n"_1"n"_2 = lambda("h"/"c")`

Now, the lines are perpendicular if

`"l"_1"l"_2 + "m"_1"m"_2 + "n"_1"n"_2 = 0`

i.e. if `lambda ("f"/"a") + lambda("g"/"b") + lambda("h"/"c") = 0`

i.e. if `"f"/"a" + "g"/"b" + "h"/"c" = 0`