    Question

# lf a1,b1,c1 and a2,b2,c2 are the d.r.s of two lines then (a1a2+b1b2+c1c2)2+Σ(bIc2−b2C1)2=

A
(a1+b1+c1)(a2+b2+c2)
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B
(a1b1c1)(a2b2C2)
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C
1
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D
(a21+b21+c21)(a22+b22+c22)
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Solution

## The correct option is D (a21+b21+c21)(a22+b22+c22)Given a1,b1,c1 and a2,b2,c2 are d.r.s of two linesVector equation of lines with a1,b1,c1 as d.r.s isa1^i+b1^j+c1^kVector equation of line with a2,b2,c2 as d.r.s is a2^i+b2^j+c2^kLet θ be the angle between two linesFormula: Dot product A⋅B=|A||B|cosθCross product: |A×B|=|A||B|sinθDot product of two linesa1a2+b1b2+c1c2=(√a12+b12+c12)(√a22+b22+c22)cosθSquaring on both sides(a1a2+b1b2+c1c2)2=(a12+b12+c12)(a22+b22+c22)cos2θcos2θ=(a1a2+b1b2+c1c2)2(a12+b12+c12)(a22+b22+c22)...(1)Cross ProductA×B=∣∣ ∣ ∣∣^i^j^ka1b1c1a2b2c2∣∣ ∣ ∣∣A×B=^i(b1c2−b2c1)+^j(c1a2−c2a1)+^k(a1b2−a2b1)|A×B|=√(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2|A×B|=√∑(b1c2−b2c1)2By cross product formula,√∑(b1c2−b2c1)2=(√a12+b12+c12)(√a22+b22+c22)sinθSquaring on both sides,∑(b1c2−b2c1)2=(a12+b12+c12)(a22+b22+c22)sin2θsin2θ=∑(b1c2−b2c1)2(a12+b12+c12)(a22+b22+c22)....(2)We know thatsin2θ+cos2θ=1....(3)Adding (1),(2)(a1a2+b1b2+c1c2)2(a12+b12+c12)(a22+b22+c22)+∑(b1c2−b2c1)2(a12+b12+c12)(a22+b22+c22)=sin2θ+cos2θFrom (3),(a1a2+b1b2+c1c2)2+∑(b1c2−b2c1)2=(a12+b12+c12)(a22+b22+c22)  Suggest Corrections  0      Similar questions  Related Videos   Property 1
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