Question
What is the angle between vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes 2 and $\sqrt{3}$ respectively? Given $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}.$
✓
Answer
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Given that
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=\sqrt{3}$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{3}=(2)(\sqrt{3})\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{3}}{2\sqrt{3}}$
$\Rightarrow\cos\theta=\frac{1}2{}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{1}2{}\big)=\frac{\pi}{3}$
Given that
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=\sqrt{3}$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{3}=(2)(\sqrt{3})\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{3}}{2\sqrt{3}}$
$\Rightarrow\cos\theta=\frac{1}2{}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{1}2{}\big)=\frac{\pi}{3}$
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