The plane 2x - 3y + 6z - 11 = 0 makes an angl | vedclass

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(C) The normal vector to the plane $2x - 3y + 6z - 11 = 0$ is $\vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.The direction vector of the $X$-axis is $\vec

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$\frac{2}{7}$

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(C) The normal vector to the plane $2x - 3y + 6z - 11 = 0$ is $\vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.The direction vector of the $X$-axis is $\vec{d} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.The angle $	heta$ between a line with direction vector $\vec{d}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{n} \cdot \vec{d}|}{|\vec{n}| |\vec{d}|}$.Calculating the dot product: $\vec{n} \cdot \vec{d} = (2)(1) + (-3)(0) + (6)(0) = 2$.Calculating the magnitudes: $|\vec{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ and $|\vec{d}| = 1$.Thus,$\sin 	heta = \frac{2}{7 	imes 1} = \frac{2}{7}$.Given that $	heta = \sin^{-1}(\alpha)$,we have $\sin^{-1}(\alpha) = \sin^{-1}(\frac{2}{7})$.Therefore,$\alpha = \frac{2}{7}$.

The plane $2x - 3y + 6z - 11 = 0$ makes an angle $\sin^{-1}(\alpha)$ with the $X$-axis. The value of $\alpha$ is equal to:

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