Two vectors A→ and B→ have equal magnitudes. If magnitude of A→+B→ is equal to four times the magnitude of A→−B→, then the angle between A→ and B→ will be :

|A→+B→|=2|A→−B→| −−−−−−(1)Given A=Bsquareing equation(1) both side.|A→+B→|2=(4|A→+B→|)2⇒A2+B2+2A→.B→=16(A2+B2−2A→.B→)⇒2A2+2A2cosθ=16(2A2−2A2cosθ)A2+A2cosθ=16A2−16A2cosθ17A2cosθ=15A2cosθ=1517

Two vectors A→ and B→ have equal magnitudes. If magnitude of A→+B→ is equal to four times the magnitude of A→−B→, then the angle between A→ and B→ will be :

|A→+B→|=2|A→−B→| −−−−−−(1)Given A=Bsquareing equation(1) both side.|A→+B→|2=(4|A→+B→|)2⇒A2+B2+2A→.B→=16(A2+B2−2A→.B→)⇒2A2+2A2cosθ=16(2A2−2A2cosθ)A2+A2cosθ=16A2−16A2cosθ17A2cosθ=15A2cosθ=1517

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Two vectors $\vec{A} a n d \vec{B}$ have equal magnitudes. If magnitude of $\vec{A} + \vec{B}$ is equal to four times the magnitude of $\vec{A} - \vec{B}$ , then the angle between $\vec{A} a n d \vec{B}$ will be :

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$\sin^{- 1} (\frac{3}{5})$

$\sin^{- 1} (\frac{4}{5})$

$\cos^{- 1} (\frac{17}{15})$

$\cos^{- 1} (\frac{15}{17})$

answer is D.

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Detailed Solution

$\begin{array}{l} | \vec{A} + \vec{B} | = 2 | \vec{A} - \vec{B} | - - - - - - (1) \\ G i v e n A = B \\ s q u a r e i n g e q u a t i o n (1) b o t h s i d e . \\ {| \vec{A} + \vec{B} |}^{2} = {(4 | \vec{A} + \vec{B} |)}^{2} \\ \Rightarrow A^{2} + B^{2} + 2 \vec{A} . \vec{B} = 16 (A^{2} + B^{2} - 2 \vec{A} . \vec{B}) \\ \Rightarrow 2 A^{2} + 2 A^{2} \cos θ = 16 (2 A^{2} - 2 A^{2} \cos θ) \\ A^{2} + A^{2} \cos θ = 16 A^{2} - 16 A^{2} \cos θ \\ 17 A^{2} \cos θ = 15 A^{2} \\ \cos θ = \frac{15}{17} \end{array}$