#### If l1, m1, n1; l2, m2, n2; l3, m3, n3 are the direction cosines of 3 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 the direction vector of the 3 mutually perpendicular lines be

$\vec{a}=l_{1}\hat{i}+m_{1}\hat{j}+n_{1}\hat{k}\\ \\ \vec{b}=l_{2}\hat{i}+m_{2}\hat{j}+n_{2}\hat{k}\\ \\ \vec{c}=l_{3}\hat{i}+m_{3}\hat{j}+n_{3}\hat{k}$

Let the direction vectors associated with direction cosines $l_{1} + l_{2} + l_{3} , m_{1} + m_{2} + m_{3} , n_{1} + n_{2} + n_{3}$ be

$\vec{p}=\left (l_{1} + l_{2} + l_{3} \right )\hat{i}+\left (m_{1} + m_{2} + m_{3} \right )\hat{j}+\left ( n_{1} + n_{2} + n_{3} \right )\hat{k}$

Since the lines associated with the direction vectors a, b, and c are mutually perpendicular, we get

$\vec{a}.\vec{b}=0$ (Since the dot product of two perpendicular vectors is 0)

=>$l_{1}l_{2} + m_{1}m_{2}+n_{1}n_{2}=0$ …(1)

Similarly,

$l_{1}l_{3} + m_{1}m_{3}+n_{1}n_{3}=0$ …(2)

Finally, $\vec{b}.\vec{c}=0$

=>$l_{2}l_{3} + m_{2}m_{3}+n_{2}n_{3}=0$ …(3)

Now, let us consider x, y and z as the angles made by direction vectors a, b, and c respectively with p.

Then,

$\cos x=\vec{a}.\vec{p}$
$\Rightarrow \cos x =l_{1}\left ( l_{1}+l_{2}+l_{3} \right )+m_{1}\left ( m_{1}+m_{2}+m_{3} \right )+n_{1}\left ( n_{1}+n_{2}+n_{3} \right )\\ \Rightarrow \cos x=l_{1}^{2}+l_{1}l_{2}+l_{1}l_{3}+m_{1}^{2}+m_{1}m_{2}+m_{1}m_{3}+n_{1}^{2}+n_{1}n_{2}+n_{1}n_{3}\\ \Rightarrow cos x= l_{1}^{2}+m_{1}^{2}+n_{1}^{2}+\left ( l_{1}l_{2}+m_{1}m_{2}+ n_{1}n_{2} \right )+\left ( l_{1}l_{3}+m_{1}m_{3}+ n_{1}n_{3} \right )\\$

We know, $l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=1$  [since the sum of squares of direction cosines of a line = 1]

$\Rightarrow \cos x=1+0=1$ [from (1) and (2)]

Then, $\cos y=1$ and $\cos z=1$

=> x = y = z = 0.

Therefore, the vector p makes equal angles with the vectors a, b and c.