Question

Find the angle between the lines $\vec{r} = (2 \hat{i} - 5 \hat{j} + \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$ and $\vec{r} = (7 \hat{i} - 6 \hat{k}) + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})$

Accepted Answer

Given lines are $\vec{r} = (2 \hat{i} - 5 \hat{j} + \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$ and $\vec{r} = (7 \hat{i} - 6 \hat{k}) + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})$

Direction ratios of lines are $3, 2, 6$ and $1, 2, 2$ respectively

The angle between the lines with direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ is given by $\cos θ = |\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}}|$

Hence angle between given lines is $θ = \cos^{- 1} |\frac{3 (1) + 2 (2) + 6 (2)}{\sqrt{3^{2} + 2^{2} + 6^{2}} \sqrt{1^{2} + 2^{2} + 2^{2}}}|$

$\Rightarrow θ = \cos^{- 1} |\frac{3 + 4 + 12}{\sqrt{9 + 4 + 36} \sqrt{1 + 4 + 4}}|$

$= \cos^{- 1} \frac{19}{7 \times 3}$

$\Rightarrow θ = \cos^{- 1} \frac{19}{21}$

Hence, the angle between the lines $\vec{r} = (2 \hat{i} - 5 \hat{j} + \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$ and $\vec{r} = (7 \hat{i} - 6 \hat{k}) + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ is $\cos^{- 1} \frac{19}{21}$

R S Aggarwal Solutions for Exercise 6: EXERCISE

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 6: EXERCISE

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