The angle between the line bar{r}=(hat{i}+2ha | vedclass

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(C) The angle $	heta$ between the line $\bar{r}=\bar{a}+\lambda\bar{b}$ and the plane $\bar{r} \cdot \bar{n}=p$ is given by $\sin 	heta = \frac{|\ba

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$

Vedclass · Answer

(C) The angle $	heta$ between the line $\bar{r}=\bar{a}+\lambda\bar{b}$ and the plane $\bar{r} \cdot \bar{n}=p$ is given by $\sin 	heta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}| |\bar{n}|}$.Here,$\bar{b} = \hat{i} - \hat{j} + \hat{k}$ and $\bar{n} = 2\hat{i} - \hat{j} + \hat{k}$.Calculating the dot product: $\bar{b} \cdot \bar{n} = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4$.Calculating the magnitudes: $|\bar{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|\bar{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.Substituting these values into the formula: $\sin 	heta = \frac{4}{\sqrt{3} \cdot \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.Therefore,$	heta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}ight)$.

The angle between the line $\bar{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot (2\hat{i}-\hat{j}+\hat{k})=4$ is:

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