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

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Question

(A) The direction vectors of the two lines are $\vec{b_1} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b_2} = 2\hat{i} + \hat{j} - \hat{k}$.Let $	heta$

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$\cos^{-1}\left(\frac{1}{6}\right)$

Vedclass · Answer

(A) The direction vectors of the two lines are $\vec{b_1} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b_2} = 2\hat{i} + \hat{j} - \hat{k}$.Let $	heta$ be the angle between the lines.The formula for the angle between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by $\cos 	heta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(2) + (1)(1) + (2)(-1) = 2 + 1 - 2 = 1$.Calculating the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$ and $|\vec{b_2}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6}$.Substituting these values into the formula: $\cos 	heta = \frac{|1|}{\sqrt{6} \cdot \sqrt{6}} = \frac{1}{6}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{6}ight)$.

The angle between the lines $\bar{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}+\hat{j}+2\hat{k})$ and $\bar{r}=(3\hat{i}+\hat{k})+\lambda^{\prime}(2\hat{i}+\hat{j}-\hat{k})$,where $\lambda, \lambda^{\prime} \in R$,is

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