Answer :

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)


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