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Question

(D) The transformation equations for rotating the axes by an angle $\alpha$ are given by:$x' = x \cos \alpha + y \sin \alpha$$y' = -x \sin \alpha + y

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(D) The transformation equations for rotating the axes by an angle $\alpha$ are given by:$x' = x \cos \alpha + y \sin \alpha$$y' = -x \sin \alpha + y \cos \alpha$Given $(x, y) = (1, 2)$ and $(x', y') = \left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}, \frac{\sqrt{3}+3}{2 \sqrt{2}}\right)$:$\frac{3 \sqrt{3}-1}{2 \sqrt{2}} = 1 \cos \alpha + 2 \sin \alpha$  $(1)$$\frac{\sqrt{3}+3}{2 \sqrt{2}} = -1 \sin \alpha + 2 \cos \alpha$  $(2)$Multiply $(1)$ by $2$ and $(2)$ by $1$ and add them:$2 \left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}\right) + \frac{\sqrt{3}+3}{2 \sqrt{2}} = 2 \cos \alpha + 4 \sin \alpha - \sin \alpha + 2 \cos \alpha$$\frac{6 \sqrt{3}-2 + \sqrt{3}+3}{2 \sqrt{2}} = 4 \cos \alpha + 3 \sin \alpha$ (This approach is complex,let's use $x'^2 + y'^2 = x^2 + y^2$)$x'^2 + y'^2 = 1^2 + 2^2 = 5$$\left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}\right)^2 + \left(\frac{\sqrt{3}+3}{2 \sqrt{2}}\right)^2 = \frac{27+1-6 \sqrt{3}}{8} + \frac{3+9+6 \sqrt{3}}{8} = \frac{28+12}{8} = 5$ (Consistent)From $(1)$ and $(2)$,solving for $\alpha$ gives $\alpha = 15^{\circ} = \frac{\pi}{12}$.

By rotating the coordinate axes in the positive direction about the origin by an angle $\alpha$,if the point $(1,2)$ is transformed to $\left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}, \frac{\sqrt{3}+3}{2 \sqrt{2}}\right)$ in the new coordinate system,then $\alpha=$

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