If l₁, m₁, n₁; l₂, m₂, n₂; l₃, m₃, n₃ are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l₁ + l₂ + l₃, m₁ + m₂ + m₃, n₁ + n₂ + n₃ makes equal angles with them.

Let direction vector of three mutually perpendicular lines be

Let the direction vector associated with directions cosines l₁ + l₂ + l₃, m₁ + m₂ + m₃, n₁ + n₂ + n₃ be

As, lines associated with direction vectors a, b and c are mutually perpendicular, we have

[dot product of two perpendicular vector is 0]

⇒ l₁l₂ + m₁m₂ + n₁n₂ = 0 [1]

Similarly,

⇒ l₁l₃ + m₁m₃ + n₁n₃ = 0 [2]

And

⇒l₂l₃ + m₂m₃ + n₂n₃ = 0 [3]

Now, let x, y, z be the angles made by direction vectors a, b and c respectively with p

Therefore,

⇒ cos x = l₁(l₁ + l₂ + l₃) + m₁(m₁ + m₂ + m₃) + n₁(n₁+ n₂ + n₃)

⇒ cos x = l₁² + l₁l₂ + l₁l₃ + m₁² + m₁m₂ + m₁m₃ + n₁² + n₁n₂ + n₁n₃

⇒ cos x = l₁² + m₁² + n₁² + (l₁l₂ + m₁m₂ + n₁n₂) + (l₁l₃ + m₁m₃ + n₁n₃)

As we know l₁² + m₁² + n₁² = 1 because sum of squares of direction cosines of a line is equal to 1

⇒ cos x = 1 + 0 = 1 [From, 1 and 2]

Similarly, cos y = 1 and cos z = 1

⇒ x = y = z = 0

Hence, angle made by vector p, with vectors a, b and c are equal!