Find the angle between the line vec{r} = (2ha | vedclass

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(C) The line is parallel to the vector $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$.The plane has a normal vector $\vec{n} = 3\hat{i} + 2\hat{j} - \hat{k}

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$\sin^{-1}\left(\frac{2}{\sqrt{42}}\right)$

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(C) The line is parallel to the vector $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$.The plane has a normal vector $\vec{n} = 3\hat{i} + 2\hat{j} - \hat{k}$.Let $	heta$ be the angle between the line and the plane. The formula for the angle is given by $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Calculating the dot product: $\vec{b} \cdot \vec{n} = (-1)(3) + (1)(2) + (1)(-1) = -3 + 2 - 1 = -2$.Calculating the magnitudes: $|\vec{b}| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{n}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$.Thus,$\sin 	heta = \frac{|-2|}{\sqrt{3} \cdot \sqrt{14}} = \frac{2}{\sqrt{42}}$.Therefore,$	heta = \sin^{-1}\left(\frac{2}{\sqrt{42}}ight)$.

Find the angle between the line $\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(-\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 4$.

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