If $∣ ∣ ˆ a + ˆ b ∣ ∣ = \sqrt 3$ $|hata+hatb| =√3$ then $∣ ∣ ˆ a - ˆ b ∣ ∣$ $|hata -hat b|$ is?

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If $∣ ∣ ˆ a + ˆ b ∣ ∣ = \sqrt 3$ $|hata+hatb| =√3$ then $∣ ∣ ˆ a - ˆ b ∣ ∣$ $|hata -hat b|$ is?

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Answer 1 · 2017-05-25T11:20:32

$∣ ∣ ˆ a + ˆ b ∣ ∣ = \sqrt 3$ $|hata+hatb| =√3$

$\Rightarrow \sqrt{{| ˆ a |}^{2} + {∣ ∣ ˆ b ∣ ∣}^{2} + 2 | ˆ a | ∣ ∣ ˆ b ∣ ∣ cos θ} = \sqrt{3}$ $=>sqrt(abshata^2+abshatb^2+2abshataabshatbcostheta)=sqrt3$

$\Rightarrow \sqrt{1^{2} + 1^{2} + 2 \times 1 \times 1 cos θ} = \sqrt{3}$ $=>sqrt(1^2+1^2+2xx1xx1costheta)=sqrt3$

[ $| ˆ a | = 1 and ∣ ∣ ˆ b ∣ ∣ = 1$ $abshata=1 and abs hatb=1$ as they are unit vector]

$\Rightarrow 1^{2} + 1^{2} + 2 \times 1 \times 1 cos θ = 3$ $=>1^2+1^2+2xx1xx1costheta=3$

$\Rightarrow cos θ = \frac{3 - 2}{2} = \frac{1}{2} = {cos 60}^{\circ}$ $=>costheta=(3-2)/2=1/2=cos60^@$

$\Rightarrow θ = 60^{\circ}$ $=>theta=60^@$

Now

$∣ ∣ ˆ a - ˆ b ∣ ∣$ $|hata -hat b|$

$= \sqrt{{| ˆ a |}^{2} + {∣ ∣ ˆ b ∣ ∣}^{2} + 2 | ˆ a | ∣ ∣ ˆ b ∣ ∣ cos (π - θ)}$ $=sqrt(abshata^2+abshatb^2+2abshataabshatbcos(pi-theta))$

$\Rightarrow \sqrt{1^{2} + 1^{2} - 2 \times 1 \times 1 cos θ}$ $=>sqrt(1^2+1^2-2xx1xx1costheta)$

$\Rightarrow \sqrt{1^{2} + 1^{2} - 2 \times 1 \times 1 {cos 60}^{\circ}}$ $=>sqrt(1^2+1^2-2xx1xx1cos60^@$

$\Rightarrow \sqrt{1^{2} + 1^{2} - 2 \times 1 \times 1 \times \frac{1}{2}} = 1$ $=>sqrt(1^2+1^2-2xx1xx1xx1/2)=1$

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If $∣ ∣ ˆ a + ˆ b ∣ ∣ = \sqrt 3$ $|hata+hatb| =√3$ then $∣ ∣ ˆ a - ˆ b ∣ ∣$ $|hata -hat b|$ is?

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If ∣∣ˆa+ˆb∣∣=√3|hata+hatb| =√3 then ∣∣ˆa−ˆb∣∣|hata -hat b| is?

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If $∣ ∣ ˆ a + ˆ b ∣ ∣ = \sqrt 3$ $|hata+hatb| =√3$ then $∣ ∣ ˆ a - ˆ b ∣ ∣$ $|hata -hat b|$ is?