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Question
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.
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Solution
Let direction vector of three mutually perpendicular lines be
`veca = l_1hati + m_1hatj + n_1hatk`
`vecb = l_2hati + m_2hatj + n_2hatk`
`vecc = l_3hati + m_3hatj + n_3hatk`
Let the direction vector associated with directions cosines l1 + l2 + l3, m1 + m2 + m3, n1 + n2 + n3 be
`vecp = (1_1 + l_2 + l_3)hati + (m_1 + m_2 + m_3)hati + (n_1 + n_2 + n_3)hatk`
As, lines associated with direction vectors a, b and c are mutually perpendicular
We have `veca . vecb` = 0 ......[dot product of two perpendicular vector is 0]
⇒ l1l2 + m1m2 + n1n2 = 0 .....[1]
Similarly, `veca . vecc` = 0
⇒ l1l3 + m1m3 + n1n3 = 0 ......[2]
And `vecb . vecc` = 0
⇒ 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, `cosx = veca. vecp`
⇒ 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!
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