Question

Find the angle between the lines whose direction ratios are proportional to a, b, c and b − c, c − a, a − b.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathit{}\theta \mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{given}\mathrm{lines}. \\ \mathrm{We}\mathrm{have} \\ {a}_1=a,b_1=b,c_1=c \\ {a}_2=b-c,b_2=c-a,c_2=a-b \\ \mathrm{Now}, \\ \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}} \\ =\frac{a\left(b-c\right)+b\left(c-a\right)+c\left(a-b\right)}{\sqrt{a^2+b^2+c^2}\sqrt{{\left(b-c\right)}^2+\left(c-a\right)\left(a-b\right)}}=\frac{ab-ac+bc-ab+ac-bc}{\sqrt{a^2+b^2+c^2}\sqrt{{\left(b-c\right)}^2+\left(c-a\right)\left(a-b\right)}}=0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \\ \mathrm{Thus},\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{measures}90°. \end{gathered}$$