If the angle between the planes x-2y+3z-5=0 a | vedclass

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(B) The normal vectors to the planes are $\vec{n_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{n_2} = \hat{i} + \alpha\hat{j} + 2\hat{k}$.The angle $\

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$\frac{62}{55}$

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(B) The normal vectors to the planes are $\vec{n_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{n_2} = \hat{i} + \alpha\hat{j} + 2\hat{k}$.The angle $	heta$ between the planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.Given $\cos 	heta = \frac{1}{14}$,we have $\frac{|(1)(1) + (-2)(\alpha) + (3)(2)|}{\sqrt{1^2 + (-2)^2 + 3^2} \sqrt{1^2 + \alpha^2 + 2^2}} = \frac{1}{14}$.$\frac{|7 - 2\alpha|}{\sqrt{14} \sqrt{\alpha^2 + 5}} = \frac{1}{14}$.Squaring both sides: $\frac{(7 - 2\alpha)^2}{14(\alpha^2 + 5)} = \frac{1}{196}$.$\frac{49 - 28\alpha + 4\alpha^2}{\alpha^2 + 5} = \frac{14}{196} = \frac{1}{14}$.$14(4\alpha^2 - 28\alpha + 49) = \alpha^2 + 5$.$56\alpha^2 - 392\alpha + 686 = \alpha^2 + 5$.$55\alpha^2 - 392\alpha + 681 = 0$.Let the roots be $\alpha_1$ and $\alpha_2$. The difference is $|\alpha_1 - \alpha_2| = \frac{\sqrt{D}}{|a|} = \frac{\sqrt{(-392)^2 - 4(55)(681)}}{55}$.$D = 153664 - 149820 = 3844$.$\sqrt{D} = \sqrt{3844} = 62$.Difference $= \frac{62}{55}$.

If the angle between the planes $x-2y+3z-5=0$ and $x+\alpha y+2z+7=0$ is $\cos^{-1}\left(\frac{1}{14}\right)$,then the difference between the values of $\alpha$ is

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