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-

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

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

Commented by Tinkutara last updated on 05/Jun/17

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

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