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If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are (m1n2 - m2n1), (n1l2 - n2l1), (l1m2 - l2m1)
Let l, m, n be the direction cosines of the line perpendicular to each of the given lines. Then,
ll1 + mm1 + nn1 = 0 …(1)
and, ll2 + mm2 + nn2 = 0 …(2)
On solving (1) and (2) by cross - multiplication, we get -
Thus, the direction cosines of the given line are proportional to
(m1n2 - m2n1), (n1l2 - n2l1), (l1m2 - l2m1)
So, its direction cosines are
where .
we know that -
(l12 + m12 + n12) (l22 + m22 + n22) - (l1l2 + m1m2 + n1n2)2
= (m1n2 - m2n1)2 + (n1l2 - n2l1)2 + (l1m2 - l2m1)2 …(3)
It is given that the given lines are perpendicular to each other. Therefore,
l1l2 + m1m2 + n1n2 = 0
Also, we have
l12 + m12 + n12 = 1
and, l22 + m22 + n22 = 1
Putting these values in (3), we get -
(m1n2 - m2n1)2 + (n1l2 - n2l1)2 + (l1m2 - l2m1)2 = 1
⇒ λ = 1
Hence, the direction cosines of the given line are (m1n2 - m2n1), (n1l2 - n2l1), (l1m2 - l2m1)