Question

Find acute & obtuse angle between two line $1$ has $D.C^{\prime }s\equiv 2/7,3/7,6/7$ & line $2$ has $D.C^{\prime }s\equiv 3/5\sqrt{2},4/5\sqrt{2},5/5\sqrt{2}$

Answer 1

According to question.............

We have,

$\begin{array}{c}\frac{2}{7},\frac{3}{7},\frac{6}{7}--------\left(1\right)\\ \frac{3}{5\sqrt{2}},\frac{4}{5\sqrt{2}},\frac{5}{5\sqrt{2}}------\left(2\right)\\ so,\\ co{s}^{-1}\theta =\left|\frac{\frac{6}{35\sqrt{2}}+\frac{12}{35\sqrt{2}}+\frac{30}{35\sqrt{2}}}{\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}\sqrt{\frac{9}{50}+\frac{16}{50}+\frac{25}{50}}}\right|\\ =\left|\frac{\frac{48}{35\sqrt{2}}}{\sqrt{\frac{49}{49}}\sqrt{\frac{50}{50}}}\right|\\ =\left|\frac{\frac{48}{35\sqrt{2}}}{\sqrt{1}\sqrt{1}}\right|=\frac{48}{35\sqrt{2}}\times \frac{1}{2}=\frac{24}{35\sqrt{2}}=0.48\\ {\cos }^{-1}\theta =0.48\\ \theta ={61}^0.31\\ \\ \therefore accute\left(below{90}^0\right)={61}^0.31\\ obtuse\left(above{90}^{0}below{180}^0\right)={180}^0-{61}^0.31\\ ={118}^0.69\end{array}$