Question

The angle between the lines whose direction ratios are proportional to a, b, c and $\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}$ is _________________.

Answer 1

Since, the angle between the lines whose direction ratios are proportional to (a₁, b₁, c₁) and (a₂, b₂, c₂)
is $\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

∴ angle between the lines whose direction ratios are proportional to

$\left(a,b,c\right)\mathrm{and}\left(\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\right)$

$\begin{array}{rcl}\mathrm{is}\cos \theta & =& \frac{a\left({\displaystyle \frac{1}{bc}}\right)+b\left({\displaystyle \frac{1}{ca}}\right)+c\left({\displaystyle \frac{1}{ab}}\right)}{\sqrt{a^2+b^2+c^2}\sqrt{{\displaystyle \frac{1}{b^2{c}^2}}+{\displaystyle \frac{1}{c^2{a}^2}}+{\displaystyle \frac{1}{a^2{b}^2}}}}\\ & =& \frac{{\displaystyle \frac{a}{bc}}+{\displaystyle \frac{b}{ca}}+{\displaystyle \frac{c}{ab}}}{\sqrt{a^2+b^2+c^2}\sqrt{{\displaystyle \frac{1}{b^2{c}^2}}+{\displaystyle \frac{1}{c^2{a}^2}}+{\displaystyle \frac{1}{a^2{b}^2}}}}\\ & =& \frac{abc\left({\displaystyle \frac{a}{bc}}+{\displaystyle \frac{b}{ca}}+{\displaystyle \frac{c}{ab}}\right)}{abc\sqrt{a^2+b^2+c^2}\sqrt{{\displaystyle \frac{1}{b^2{c}^2}}+{\displaystyle \frac{1}{c^2{a}^2}}+{\displaystyle \frac{1}{a^2{b}^2}}}}\\ & =& \frac{a^2+b^2+c^2}{\sqrt{a^2+b^2+c^2}\left(\sqrt{{\displaystyle \frac{a^2{b}^2{c}^2}{b^2{c}^2}}+{\displaystyle \frac{a^2{b}^2{c}^2}{c^2{a}^2}}+{\displaystyle \frac{a^2{b}^2{c}^2}{a^2{b}^2}}}\right)}\\ & =& \frac{a^2+b^2+c^2}{\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}}\\ & =& \frac{a^2+b^2+c^2}{a^2+b^2+c^2}\\ & =& 1\\ \mathrm{i}.\mathrm{e}.\cos \theta & =& 1\\ \mathrm{i}.\mathrm{e}.\theta & =& 0°\end{array}$