Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The lines are $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\text{ and }\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ Comparing equations with $\frac{x-x_1}{{\mathrm{a}}_1}=\frac{y-y_1}{{\mathrm{b}}_1}=\frac{\mathrm{z}-{\mathrm{z}}_1}{{\mathrm{c}}_1}\text{ and }\frac{x-x_2}{{\mathrm{a}}_2}=\frac{y-y_2}{{\mathrm{b}}_2}=\frac{\mathrm{z}-{\mathrm{z}}_2}{{\mathrm{c}}_2}\text{, }$ we get $\begin{array}{rl}& x_1=1,y_1=2,{\mathrm{z}}_1=3x_2=2,y_2=4,{\mathrm{z}}_2=5\\ & {\mathrm{a}}_1=2,{\mathrm{b}}_1=3,{\mathrm{c}}_1=4{\mathrm{a}}_2=3,{\mathrm{b}}_2=4,{\mathrm{c}}_2=5\\ & \left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ {\mathrm{a}}_1& {\mathrm{b}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\end{array}\right|=\left|\begin{array}{ccc}1& 2& 2\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|\\ & =1(15-16)-2(10-12)+2(8-9)\\ & =1\\ & \therefore \sqrt{{\left(a_1{b}_2-a_2{b}_1\right)}^2+{\left(b_1{c}_2-b_2{c}_1\right)}^2+{\left(c_1{a}_2-c_2{a}_1\right)}^2}\\ & =\sqrt{(2\times 4-3\times 3{)}^2+(3\times 5-4\times 4{)}^2+(4\times 3-5\times 2{)}^2}\\ & =\sqrt{1+1+4}\\ & =\sqrt{6}\\ & \end{array}$ Shortest distance between line is $\mathrm{d}$ $\begin{array}{rl}\mathrm{d}& =\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{a}}_1& {\mathrm{b}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\end{array}\right|}{\sqrt{{\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)}^2+{\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1\right)}^2+\sqrt{{\left({\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{c}}_2{\mathrm{a}}_1\right)}^2}}}\\ & =\frac{1}{\sqrt{6}}\end{array}$