Find the angle between the plane bar{r} cdot | vedclass

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Question

(A) The direction vector of the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-1}$ is $\bar{l} = (2, 1, -1)$.The normal vector to the plane $\bar{r} \cdo

Vedclass · Accepted Answer

$\pi /6$

Vedclass · Answer

(A) The direction vector of the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-1}$ is $\bar{l} = (2, 1, -1)$.The normal vector to the plane $\bar{r} \cdot (1, 2, 1) = 1$ is $\bar{n} = (1, 2, 1)$.The angle $\alpha$ between a line and a plane is given by $\sin \alpha = \frac{|\bar{l} \cdot \bar{n}|}{|\bar{l}| |\bar{n}|}$.Calculating the dot product: $\bar{l} \cdot \bar{n} = (2)(1) + (1)(2) + (-1)(1) = 2 + 2 - 1 = 3$.Calculating the magnitudes: $|\bar{l}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\bar{n}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.Substituting these values: $\sin \alpha = \frac{|3|}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.Therefore,$\alpha = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$.

Find the angle between the plane $\bar{r} \cdot (1, 2, 1) = 1$ and the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-1}$.

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