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Question Number 14882 by Tinkutara last updated on 05/Jun/17

Two vectors a^→ and b^→ are parallel and have same magnitude. Then they (1) have same direction, but they are not equal (2) are equal (3) are not equal (4) may or may not be equal

$$\mathrm{Two}\:\mathrm{vectors}\:\overset{\rightarrow} {{a}}\:\mathrm{and}\:\overset{\rightarrow} {{b}}\:\mathrm{are}\:\mathrm{parallel}\:\mathrm{and} \\ $$$$\mathrm{have}\:\mathrm{same}\:\mathrm{magnitude}.\:\mathrm{Then}\:\mathrm{they} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{have}\:\mathrm{same}\:\mathrm{direction},\:\mathrm{but}\:\mathrm{they}\:\mathrm{are} \\ $$$$\mathrm{not}\:\mathrm{equal} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{are}\:\mathrm{equal} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{are}\:\mathrm{not}\:\mathrm{equal} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{may}\:\mathrm{or}\:\mathrm{may}\:\mathrm{not}\:\mathrm{be}\:\mathrm{equal} \\ $$

Commented by Tinkutara last updated on 05/Jun/17

Answer is (2). But how?

$$\mathrm{Answer}\:\mathrm{is}\:\left(\mathrm{2}\right).\:\mathrm{But}\:\mathrm{how}? \\ $$

Commented by ajfour last updated on 05/Jun/17

a^� =r(sin φcos θ i^� +sin φsin θ j^� +cos φ k^� ) b^� =r(sin φcos θ i^� +sin φsin θ j^� +cos φ k^� ) so , a^� =b^� ; (r, θ, φ) cannot be compensated for .

$$\bar {{a}}={r}\left(\mathrm{sin}\:\phi\mathrm{cos}\:\theta\:\hat {{i}}+\mathrm{sin}\:\phi\mathrm{sin}\:\theta\:\hat {{j}}+\mathrm{cos}\:\phi\:\hat {{k}}\right) \\ $$$$\bar {{b}}={r}\left(\mathrm{sin}\:\phi\mathrm{cos}\:\theta\:\hat {{i}}+\mathrm{sin}\:\phi\mathrm{sin}\:\theta\:\hat {{j}}+\mathrm{cos}\:\phi\:\hat {{k}}\right) \\ $$$$\:{so}\:,\:\bar {{a}}=\bar {{b}}\:\:\:\:\:\:;\:\:\:\left({r},\:\theta,\:\phi\right)\:{cannot}\:{be} \\ $$$${compensated}\:{for}\:. \\ $$

Commented by Tinkutara last updated on 05/Jun/17

But their directions can be opposite. In that case a^→ ≠ b^→ , rather a^→ = −b^→ . In this case also ∣a^→ ∣ = ∣b^→ ∣. So in my view, answer should be (4).

$$\mathrm{But}\:\mathrm{their}\:\mathrm{directions}\:\mathrm{can}\:\mathrm{be}\:\mathrm{opposite}.\:\mathrm{In} \\ $$$$\mathrm{that}\:\mathrm{case}\:\overset{\rightarrow} {{a}}\:\neq\:\overset{\rightarrow} {{b}},\:\mathrm{rather}\:\overset{\rightarrow} {{a}}\:=\:−\overset{\rightarrow} {{b}}.\:\mathrm{In}\:\mathrm{this} \\ $$$$\mathrm{case}\:\mathrm{also}\:\mid\overset{\rightarrow} {{a}}\mid\:=\:\mid\overset{\rightarrow} {{b}}\mid.\:\mathrm{So}\:\mathrm{in}\:\mathrm{my}\:\mathrm{view}, \\ $$$$\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\left(\mathrm{4}\right). \\ $$

Commented by mrW1 last updated on 05/Jun/17

the answer should be (4).

$${the}\:{answer}\:{should}\:{be}\:\left(\mathrm{4}\right). \\ $$

Commented by Tinkutara last updated on 05/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Commented by ajfour last updated on 05/Jun/17

tbere is a word antiparallel, we use it for vectors , not for lines..

$${tbere}\:{is}\:{a}\:{word}\:{antiparallel},\:{we} \\ $$$${use}\:{it}\:{for}\:{vectors}\:,\:{not}\:{for}\:{lines}.. \\ $$

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