The direction cosines of three mutually perpendicular straight lines are l1,m1,n1,l2,m2,n2,l3,m3,n3. Then the direction cosines of a line which is equally inclined to the given three lines, are

A
$(l_{1} + l_{2} + l_{3}), (m_{1} + m_{2} + m_{3}), (n_{1} + n_{2} + n_{3})$
B
$(\frac{l_{1} + l_{2} + l_{3}}{3}), (\frac{m_{1} + m_{2} + m_{3}}{3}), (\frac{n_{1} + n_{2} + n_{3}}{3})$
C
$(\frac{l_{1} + l_{2} + l_{3}}{\sqrt{3}}), (\frac{m_{1} + m_{2} + m_{3}}{\sqrt{3}}), (\frac{n_{1} + n_{2} + n_{3}}{\sqrt{3}})$
D
$(\frac{l_{1} + l_{2} + l_{3}}{\sqrt{2}}), (\frac{m_{1} + m_{2} + m_{3}}{\sqrt{2}}), (\frac{n_{1} + n_{2} + n_{3}}{\sqrt{2}})$

Solution:

Given $〈l_{1}, m_{1}, n_{1}〉, 〈l_{2}, m_{2}, n_{2}〉, 〈l_{3}, m_{3}, n_{3}〉$ are the direction cosines of three mutually perpendicular lines

It implies that $\sum l_{1} l_{2} = \sum l_{2} l_{3} = \sum l_{1} l_{3} = 0, \sum {l_{1}}^{2} = \sum {l_{2}}^{2} = \sum {l_{3}}^{2} = 1$

Hence the direction cosines of line which is equally inclined to the above lines are $(\frac{l_{1} + l_{2} + l_{3}}{3}, \frac{m_{1} + m_{2} + m_{3}}{3}, \frac{n_{1} + n_{2} + n_{3}}{3},)$

The direction cosines of three mutually perpendicular straight lines are $〈l_{1}, m_{1}, n_{1}〉, 〈l_{2}, m_{2}, n_{2}〉, 〈l_{3}, m_{3}, n_{3}〉$ . Then the direction cosines of a line which is equally inclined to the given three lines, are

Solution:

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