Q. 2

# 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)

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