If $l_{1}, m_{1}, n_{1}, l_{2}, m_{2}, n_{2} a n d l_{3}, m_{3}, n_{3}$ are the direction cosines of three mutually perpendicular lines,then prove that the line whose direction cosines are proportional to $l_{1} + l_{2} + l_{3}, m_{1} + m_{2} + m_{3} a n d n_{1} + n_{2} + n_{3}$ makes equal angles with them.

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Solution

Let $\to a = l_{1}^i + m_{1}^j + n_{1}^k \to b = l_{2}^i + m_{2}^j + n_{2}^k \to c = l_{3}^i + m_{3}^j + n_{3}^k \to d = (l_{1} + l_{2} + l_{3})^i + (m_{1} + m_{2} + m_{3})^j + (n_{1} + n_{2} + n_{3})^k A l s o, l e t α, β a n d γ a r e t h e a n g l e s b e t w e e n \to a a n d \to d, \to b a n d \to d, \to c a n d \to d . ∴ c o s α = l_{1} (l_{1} + l_{2} + l_{3}) + m_{1} (m_{1} + m_{2} + m_{3}) + n_{1} (n_{1} + n_{2} + n_{3}) = l_{1}^{2} + l_{1} l_{2} + l_{1} l_{3} + m_{1}^{2} + m_{1} m_{2} + m_{1} m_{3} + n_{2} + n_{1} n_{2} + n_{1} n_{3} = (l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) + (l_{1} l_{2} + l_{1} l_{3} + m_{1} m_{2} + m_{1} m_{3} + n_{1} n_{2} + n_{1} n_{3}) = 1 + 0 = 1 [∵ l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1 a n d l_{1} ⊥ l_{2}, l_{1} ⊥ l_{3} + m_{1} ⊥ m_{2}, m_{1} ⊥ m_{3}, n_{1} ⊥ n_{2}, n_{1} ⊥ n_{3}] S i m i l a r l y, c o s β = l_{2} (l_{1} + l_{2} + l_{3}) + m_{2} (m_{1} + m_{2} + m_{3}) + n_{2} (n_{1} + n_{2} + n_{3}) = 1 + 0 a n d c o s γ = 1 + 0 \Rightarrow c o s α = c o s β = c o s γ \Rightarrow α = β = γ$

So,the line whose direction cosines are proportional to $l_{1} + l_{2} + l_{3}, m_{1} + m_{2} + m_{3}, n_{1} + n_{2} + n_{3}$ makes equal angles with the three mutually perpendicular lines whose direction cosines are $l_{1} m_{1} n_{1}, l_{2} m_{2} n_{2}, l_{3} m_{3} n_{3},$ respectively.

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