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(A) The given lines are parallel to the vectors $\vec{b}_{1}=\hat{i}-\hat{j}-2\hat{k}$ and $\vec{b}_{2}=3\hat{i}-5\hat{j}-4\hat{k}$ respectively.First

Vedclass · Accepted Answer

$Q=\cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)$

Vedclass · Answer

(A) The given lines are parallel to the vectors $\vec{b}_{1}=\hat{i}-\hat{j}-2\hat{k}$ and $\vec{b}_{2}=3\hat{i}-5\hat{j}-4\hat{k}$ respectively.First,calculate the magnitudes of the vectors:$|\vec{b}_{1}|=\sqrt{(1)^{2}+(-1)^{2}+(-2)^{2}}=\sqrt{1+1+4}=\sqrt{6}$$|\vec{b}_{2}|=\sqrt{(3)^{2}+(-5)^{2}+(-4)^{2}}=\sqrt{9+25+16}=\sqrt{50}=5\sqrt{2}$Next,calculate the dot product $\vec{b}_{1} \cdot \vec{b}_{2}$:$\vec{b}_{1} \cdot \vec{b}_{2} = (1)(3) + (-1)(-5) + (-2)(-4) = 3 + 5 + 8 = 16$The angle $Q$ between the lines is given by:$\cos Q = \frac{|\vec{b}_{1} \cdot \vec{b}_{2}|}{|\vec{b}_{1}| |\vec{b}_{2}|}$Substituting the values:$\cos Q = \frac{16}{\sqrt{6} \cdot 5\sqrt{2}} = \frac{16}{\sqrt{2} \cdot \sqrt{3} \cdot 5\sqrt{2}} = \frac{16}{5 \cdot 2 \cdot \sqrt{3}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$Therefore,$Q = \cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)$.

Find the angle between the following pairs of lines:
$\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k})$ and
$\vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})$

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Find the angle between the following pairs of lines:$\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k})$ and$\vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})$