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Question

The direction cosines of a line satisfy the relations λ(l+m)=n and mn+nl+lm=0. The value of λ, for which the two lines are perpendicular to each other is

A
1
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B
2
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C
12
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D
3
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Solution

The correct option is A 2

Given relations between the direction cosines are λ(l+m)=n ------(1)

and lm+mn+ln=0 -------(2)

Eliminating n from (1) and (2) gives,

lm+(l+m)λ(l+m)=0

lm+λ(l2+m2+2lm)=0

Dividing equation with m2 gives,

λ(lm)2+(2λ+1)(lm)+λ=0Pproduct of roots l1l2m1m2=1 -----(3)

Now eliminating l from (1) and (2) gives,

mn+(n+m)(nλm)=0

λmn+(n+m)(nλm)=0

λm2mnn2=0

Dividign eqaution with n2 gives,,

λ(mn)2mn1=0

Product of roots m1m2n1n2=1λ ------(4)

From (3) and (4), we get

l1l21=m1m21=n1n2λ

The two lines are perpendicular to each other if l1l2+m1m2+n1n2=0

1+1λ=0

λ=2

Hence, option B is correct.


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