The angle between the line vec{r}=(hat{i}+hat | vedclass

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(C) The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left|\frac{

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$\sin^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$

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(C) The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|}ight|$.Here,the direction vector of the line is $\vec{b} = 3\hat{i} + \hat{j}$ and the normal vector to the plane is $\vec{n} = \hat{i} + 2\hat{j} + 3\hat{k}$.The dot product is $\vec{b} \cdot \vec{n} = (3)(1) + (1)(2) + (0)(3) = 3 + 2 + 0 = 5$.The magnitudes are $|\vec{b}| = \sqrt{3^2 + 1^2} = \sqrt{10}$ and $|\vec{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.Substituting these values into the formula: $\sin 	heta = \frac{5}{\sqrt{10} \cdot \sqrt{14}} = \frac{5}{\sqrt{140}} = \frac{5}{2\sqrt{35}} = \frac{\sqrt{5} \cdot \sqrt{5}}{2\sqrt{7} \cdot \sqrt{5}} = \frac{\sqrt{5}}{2\sqrt{7}}$.Thus,$	heta = \sin^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}ight)$.

The angle between the line $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3\hat{i}+\hat{j})$ and the plane $\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})=8$ is:

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