Let a vector alpha hat{i}+beta hat{j} be obta | vedclass

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Question

(A) The initial vector is $\vec{v} = \sqrt{3} \hat{i} + \hat{j}$. Its magnitude is $|\vec{v}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$. The angle

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) The initial vector is $\vec{v} = \sqrt{3} \hat{i} + \hat{j}$. Its magnitude is $|\vec{v}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$. The angle of the initial vector with the positive $x$-axis is $	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = 30^{\circ}$. Rotating this vector by $45^{\circ}$ counterclockwise gives a new vector with angle $	heta' = 30^{\circ} + 45^{\circ} = 75^{\circ}$. The new vector is $\vec{v}' = |\vec{v}|(\cos 75^{\circ} \hat{i} + \sin 75^{\circ} \hat{j}) = 2(\cos 75^{\circ} \hat{i} + \sin 75^{\circ} \hat{j})$. Thus,$\alpha = 2 \cos 75^{\circ}$ and $\beta = 2 \sin 75^{\circ}$. The vertices of the triangle are $(\alpha, \beta), (0, \beta)$,and $(0,0)$. This is a right-angled triangle with base length $|\alpha| = \alpha$ and height $|\beta| = \beta$. Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} \alpha \beta = \frac{1}{2} (2 \cos 75^{\circ}) (2 \sin 75^{\circ}) = 2 \sin 75^{\circ} \cos 75^{\circ} = \sin(2 	imes 75^{\circ}) = \sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2}$.

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