If l1,m1,n1 and 12,m2,n2 are the direction cosines of two lines, then (l1l2+m1m2+n1n2)2+∑(m1n2−m2n1)2=
A
0
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B
−1
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C
1
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D
2
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Solution
The correct option is C1 Dot product of dc's of two lines gives, (l1^i+m1^j+n1^k).(l2^i+m2^j+n2^k)=√l12+m12+n12×√l22+m22+n22×cosθ We know l12+m12+n12=1,l22+m22+n22=1 ⇒l1l2+m1m2+n1n2=1.1.cosθ⇒(l1l2+m1m2+n1n2)2=cos2θ Cross product of dc's of two lines ∣∣(l1^i+m1^j+n1^k)×(l2^i+m2^j+n2^k)∣∣=∣∣√l12+m12+n12×√l22+m22+n22×sinθ∣∣ ⇒√Σ(m1n2−m2n1)2=sinθ⇒Σ(m1n2−m2n1)2=sin2θ (l1l2+m1m2+n1n2)2+Σ(m1n2−m2n1)2=cos2θ+sin2θ=1