Find the equations of the two lines through the origin which intersect the line (x - 3) / 2 = (y -3 )/ 1 = z / 1  at angles of pi / 3 each

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}}{1}=\mathrm{t}, \mathrm{where} \mathrm{t} \mathrm{is} \mathrm{parameter}. \\ \mathrm{Direction} \mathrm{ratios} \mathrm{of} \mathrm{line} \mathrm{are}: \\ \overrightarrow{\mathrm{p}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \mathrm{Any} \mathrm{general} \mathrm{point} \mathrm{on} \mathrm{this} \mathrm{line} \mathrm{will} \mathrm{be}: \\ \left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(2\mathrm{t}+3,\mathrm{t}+3,\mathrm{t}\right) \\ \mathrm{Required} \mathrm{lines} \mathrm{will} \mathrm{pass} \mathrm{through} \left(2\mathrm{t}+3,\mathrm{t}+3,\mathrm{t}\right) \mathrm{and} \mathrm{origin}, \mathrm{hence} \mathrm{equation} \mathrm{is}: \\ \frac{\mathrm{x}-0}{2\mathrm{t}+3-0}=\frac{\mathrm{y}-0}{\mathrm{t}+3-0}=\frac{\mathrm{z}-0}{\mathrm{t}-0} \\ \Rightarrow \frac{\mathrm{x}}{2\mathrm{t}+3}=\frac{\mathrm{y}}{\mathrm{t}+3}=\frac{\mathrm{z}}{\mathrm{t}} ;\left(\mathrm{i}\right) \\ \mathrm{Direction} \mathrm{ratios} \mathrm{of} \mathrm{line} \mathrm{are}: \\ \overrightarrow{\mathrm{q}}=\left(2\mathrm{t}+3\right)\hat{\mathrm{i}}+\left(\mathrm{t}+3\right)\hat{\mathrm{j}}+\mathrm{t}\hat{\mathrm{k}} \\ \mathrm{Now} \mathrm{angle} \mathrm{between} \overrightarrow{\mathrm{p}} \mathrm{and} \overrightarrow{\mathrm{q}} \mathrm{is} \frac{\mathrm{\pi }}{3}. \\ \cos \frac{\mathrm{\pi }}{3}=\frac{\overrightarrow{\mathrm{p}}.\overrightarrow{\mathrm{q}}}{\left|\overrightarrow{\mathrm{p}}\right| \left|\overrightarrow{\mathrm{q}}\right|} \\ \Rightarrow \frac{1}{2}=\frac{\left[2\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\right].\left[\left(2\mathrm{t}+3\right)\hat{\mathrm{i}}+\left(\mathrm{t}+3\right)\hat{\mathrm{j}}+\mathrm{t}\hat{\mathrm{k}}\right]}{\sqrt{2^2+1^2+1^2}\sqrt{{\left(2\mathrm{t}+3\right)}^2+{\left(\mathrm{t}+3\right)}^2+{\mathrm{t}}^2}} \\ \Rightarrow \frac{1}{2}=\frac{2\left(2\mathrm{t}+3\right)+\left(\mathrm{t}+3\right)+\mathrm{t}}{\sqrt{4+1+1}\sqrt{4{\mathrm{t}}^2+9+12\mathrm{t}+{\mathrm{t}}^2+9+6\mathrm{t}+{\mathrm{t}}^2}} \\ \Rightarrow \frac{1}{2}=\frac{4\mathrm{t}+6+\mathrm{t}+3+\mathrm{t}}{\sqrt{6}\sqrt{6{\mathrm{t}}^2+18\mathrm{t}+18}} \\ \Rightarrow \frac{1}{2}=\frac{6\mathrm{t}+9}{\sqrt{6}\sqrt{6}\sqrt{{\mathrm{t}}^2+3\mathrm{t}+3}} \\ \Rightarrow \frac{1}{2}=\frac{6\left(\mathrm{t}+1\right)}{6\sqrt{{\mathrm{t}}^2+3\mathrm{t}+3}} \\ \Rightarrow \frac{1}{2}=\frac{3\left(2\mathrm{t}+3\right)}{6\sqrt{{\mathrm{t}}^2+3\mathrm{t}+3}} \\ \Rightarrow 1=\frac{\left(2\mathrm{t}+3\right)}{\sqrt{{\mathrm{t}}^2+3\mathrm{t}+3}} \\ \Rightarrow \sqrt{{\mathrm{t}}^2+3\mathrm{t}+3}=\left(2\mathrm{t}+3\right) \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get}: \\ \Rightarrow {\mathrm{t}}^2+3\mathrm{t}+3=4{\mathrm{t}}^2+9+12\mathrm{t} \\ \Rightarrow 3{\mathrm{t}}^2+9\mathrm{t}+6=0 \\ \Rightarrow {\mathrm{t}}^2+3\mathrm{t}+2=0 \\ \Rightarrow \left(\mathrm{t}+1\right)\left(\mathrm{t}+2\right)=0 \\ \mathrm{Hence} \\ \mathrm{t}=-1 \mathrm{or} \mathrm{t}=-2 \\ \mathrm{Put} \mathrm{the} \mathrm{obtained} \mathrm{values} \mathrm{in} \left(\mathrm{i}\right) \\ \mathrm{for} \mathrm{t}=-1 \\ \frac{\mathrm{x}}{-2+3}=\frac{\mathrm{y}}{-1+3}=\frac{\mathrm{z}}{-1} \\ \Rightarrow \frac{\mathbf{x}}{\mathbf{1}}\mathbf{=}\frac{\mathbf{y}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{z}}{\mathbf{-}\mathbf{1}}\mathbf{ } \\ \mathrm{for} \mathrm{t}=-2 \\ \frac{\mathrm{x}}{-4+3}=\frac{\mathrm{y}}{-2+3}=\frac{\mathrm{z}}{-2} \\ \Rightarrow \frac{\mathbf{x}}{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{y}}{\mathbf{1}}\mathbf{=}\frac{\mathbf{z}}{\mathbf{-}\mathbf{2}} \end{gathered}$$

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