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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.
Solution
Let `vec"a", vec"b", vec"c"` and `vec"d"` are such that
`vec"a" = l_"i"hat"i" + "m"_1hat"i" + "n"_1hat"k"`
`vec"b" = l_2hat"i" + "m"_2hat"j" + "n"_2hat"k"`
`vec"c" = l_3hat"i" + "m"_3hat"j" + "n"_3hat"k"`
And `vec"d" = (l_1 + l_2 + l_3)hat"i" + ("m"_1 + "m"_2 + "m"_3)hat"j" + ("n"_1 + "n"_2 + "n"_3)hat"k"`
Since the given d’cosines are mutually perpendicular then
l1l2 + m1m2 + n1n2 = 0
l2l3 + m2m3 + n2n3 = 0
l1l3 + m1m3 + n1n3 = 0
Let α, β and ϒ be the angles between `vec"a"` nad `vec"d"`, `vec"b"` and `vec"d"`, `vec"c"` and `vec"d"` respectively.
∴ `cos alpha = 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_1l_2 + l_1l_3 + m_1^2 + m_1m_2 + m_1m_3 + n_1^2 + n_1n_2 + n_1n_3`
= `(l_1^2 + m_1^2 + n_1^2) + (l_1l_2 + m_1m_2 + n_1n_2) + (l_1l_3 + m_1m_3 + n_1n_3)`
= 1 + 0 + 0
= 1
∴ `cos beta = l_2(l_1 + l_2 + l_3) + m_2(m_1 + m_2 + m_3) + n_2(n_1 + n_2 + n_3)`
= `l_1l_2 + l_2^2 + l_2l_3 + m_1m_2 + m_2^2 + m_2m_3 + n_1n_2 + n_2^2 + n_2n_3`
= `(l_2^2 + m_2^2 + n_2^2) + (l_1l_2 + m_1m_2 + n_1n_2) + (l_2l_3 + m_2 +m_3 + n_2n_3)`
= 1 + 0 + 0
= 1
Similarly,
∴ `cos ϒ = l_2(l_1 + l_2 + l_3) + m_2(m_1 + m_2 + m_3) + n_2(n_1 + n_2 + n_3)`
= `l_1l_3 + l_2 +l_3 + l_3^2 + m_1m-3 + m_2m_3 + m_3^2 + n_1n_3 + n_2n_3 + n_3^2`
= `(l_3^2 + m_3^2 + n_3^2) + (l_1_3 + m_1 + m_3 + n_1n_3) + (l_2l_3 + m_2 + m_3 + n_2 n_3)`
= 1 + 0 + 0
= 1
∴ `cos alpha = cos beta = cos ϒ` = 1
⇒ α = β = ϒ which is the required result.
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