Question

Find the angle between the following pairs of lines :

$r=2\hat{i}-5\hat{j}+\hat{k}+\lambda (3\hat{i}+2\hat{j}+6\hat{k})andr=7\hat{i}-6\hat{k}+\mu (\hat{i}+2\hat{j}+2\hat{k})$

$r=3\hat{i}+\hat{j}-2\hat{k}+\lambda (\hat{i}-\hat{j}-2\hat{k})andr=2\hat{i}-\hat{j}-56\hat{k}+\mu (3\hat{i}-5\hat{j}-4\hat{k})$

Answer 1

Let $\theta$ be the acute angle between the given lines, then
$cos\theta =\left|\frac{b_1.b_2}{|b_1||b_2|}\right|$
The given lines are respectively parallel to the vectors$b_1=3\hat{i}+2\hat{j}+6\hat{k}and{b}_2=\hat{i}+2\hat{j}+2\hat{k}$
$\therefore |b_1|=\sqrt{3^2+2^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7|b_2|=\sqrt{1^2+2^2+2^2}=\sqrt{1+4+4}=\sqrt{9}=3and{b}_1,b_2=(3\hat{i}+2\hat{j}+6\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})=3\times 1+2\times 2+6\times 2=3+4+12=19\therefore cos\theta =\left|\frac{b_1.b_2}{|b_1||b_2|}\right|=\frac{19}{7\times 3}=\frac{19}{21}\Rightarrow \theta =co{s}^{-1}\left(\frac{19}{21}\right)$

Let $\theta$ be the acute angle between the given lines, then
$cos\theta =\left|\frac{b_1.b_2}{|b_1||b_2|}\right|$
The given lines are respectively parallel to the vectors
$b_1=\hat{i}-\hat{j}-2\hat{k}and{b}_2=3\hat{i}-5\hat{j}-4\hat{k}$
$\therefore |b_1|=\sqrt{(1{)}^2+(-1{)}^2+(-2{)}^2}=\sqrt{1+1+4}=\sqrt{6}|b_2|=\sqrt{(3{)}^2+(-5{)}^2+(-4{)}^2}=\sqrt{9+25+16}=\sqrt{50}=5\sqrt{2}and{b}_1.b_2=(\hat{i}-\hat{j}-2\hat{k}).(3\hat{i}-5\hat{j}-4\hat{k})=1\times 3-1\times (-5)-2\times (-4)=3+5+8=16\therefore cos\theta =\left|\frac{b_1.b_2}{|b_1||b_2|}\right|=\frac{16}{\sqrt{6}\times 5\sqrt{2}}=\frac{16}{10\sqrt{3}}=\frac{8}{5\sqrt{3}}\Rightarrow \theta =co{s}^{-1}\left(\frac{8}{5\sqrt{3}}\right)$