If three mutually perpendicular lines have direction cosines $(l_{1}, m_{1}, n_{1}), (l_{2}, m_{2}, n_{2})$ and $(l_{3}, m_{3}, n_{3})$ line having direction cosines $l_{1} + l_{2} + l_{3}, m_{1} + m_{2} + m_{3}$ and $n_{1} + n_{2} + n_{3}$ make an
angle of..... with each other

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Solution

The correct option is A
0

Lines are mutually perpendicular
$∴ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0, l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} = 0$
and $l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3} = 0$
Therefore,
$θ = c o s^{- 1} [(l_{1} + l_{2} + l_{3}) l_{1} + (m_{1} + m_{2} + m_{3}) m_{1} + (n_{1} + n_{2} + n_{3}) n_{1}]$
$\Rightarrow θ = c o s^{- 1} ∣ ∣ \sum l_{1}^{2} ∣ ∣ = c o s^{- 1} (1) \Rightarrow θ = o^{\circ}$
Similarly with other lines, it will make $0^{\circ}$ angle.

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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.