# Prove that the two vectors whose direction cosines are given by relations al + bm + cn = 0 and fmn + gnl + hlm = 0 are perpendicular, if fagbhcfa+gb+hc=0 - Mathematics and Statistics

Sum

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 + bfm2 + agl2 + bglm) + chlm = 0

∴ agl2 + (af + bg - ch)lm + bfm2 = 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 l2 + m2 + n2 = 1

Let us take m ≠ 0.

Dividing equation (4) by m2, we get

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

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

If l1, m1, n1 and l2, m2, n2 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

Concept: Vector Product of Vectors (Cross)
Is there an error in this question or solution?