Answer:
Given equation of lines are \[\frac{2-x}{-\,5}=\frac{3+y}{3}=\frac{z}{2}\] \[\Rightarrow \]\[\frac{x-2}{5}=\frac{y+3}{3}=\frac{z}{2}.\] ? (i) and \[\frac{x+2}{-1}=\frac{3y-5}{2}=\frac{z-5}{4}\] \[\Rightarrow \] \[\frac{x+2}{-1}=\frac{y-(5/3)}{(2/3)}=\frac{z-5}{4}\] ? (ii) On comparing Eqs. (i) and (ii) with one point form \[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}}\] and \[\frac{x-{{x}_{2}}}{{{a}_{2}}}=\frac{y-{{y}_{2}}}{{{b}_{2}}}=\frac{z-{{z}_{2}}}{{{c}_{2}}},\] we get \[{{a}_{1}}=5,\] \[{{b}_{1}}=3,\] \[{{c}_{1}}=2\] ? (iii) \[{{a}_{2}}=-1,\] \[{{b}_{2}}=\frac{2}{3},\] \[{{c}_{2}}=4\] ? (iv) Now, we know that angle between two lines is given by \[\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}.\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] ? (v) \[\therefore \] From Eqs. (iii), (iv) and (v), we get \[\cos \theta =\frac{(5)(-1)+(3)\left( \frac{2}{3} \right)+(2)(4)}{\sqrt{25+9+4}.\sqrt{1+\frac{4}{9}+16}}\] \[\Rightarrow \] \[\cos \theta =\frac{-\,5+2+8}{\sqrt{38}\cdot \frac{\sqrt{157}}{3}}\] \[=\frac{5\times 3}{\sqrt{38}\cdot \sqrt{157}}\] \[=\frac{15}{\sqrt{5966}}\] \[\therefore \] \[\theta ={{\cos }^{-1}}\left( \frac{15}{\sqrt{5966}} \right)\]
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