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  • If the two lines <img alt="l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2 y+3}{\alpha}=\frac{z+5}{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?l_%7B1%7D%3A%

If the two lines l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2 y+3}{\alpha}=\frac{z+5}{2} are perpendicular, then an angle between the lines l_{2}$ and $l_{3}: \frac{1-x}{3}=\frac{2 y-1}{-4}=\frac{z}{4} is:

Option: 1

\begin{aligned} &\cos ^{-1}\left(\frac{29}{4}\right) \\ \end{aligned}


Option: 2

\sec ^{-1}\left(\frac{29}{4}\right) \\


Option: 3

\cos ^{-1}\left(\frac{2}{29}\right) \\


Option: 4

\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)


Answers (1)

best_answer

\mathrm{l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0} }\\

\mathrm{l_{2}: \frac{x-1}{1}=\frac{y+3 / 2}{\alpha / 2}=\frac{z+5}{2}} \\

\mathrm{l_{3}: \frac{x-1}{-3}=\frac{y-\frac{1}{2}}{-2}=\frac{z-0}{4} }

\mathrm{l_{1}\,perpendicular\,l_2: \Rightarrow \frac{|3-\alpha+0|}{\sqrt{13} \sqrt{1+\frac{\alpha^{2}}{4}}+4}=0 \Rightarrow \alpha=3}

Angle between \mathrm{l_{2}\: and\: l_{3}}

\mathrm{\cos \theta=\frac{|1 \times(-3)+(-2)(\alpha / 2)+2 \times 4 \mid}{\sqrt{1+4+\frac{\alpha^{2}}{4}} \sqrt{9+16+4}}} \\

\mathrm{\cos \theta=\frac{\mid-3-\alpha+8\mid}{\sqrt{5+\frac{\alpha^{2}}{4}} \sqrt{29}} \quad \text { put } \alpha=3}

\mathrm{\cos \theta =\frac{2}{\sqrt{\frac{29}{4}} \sqrt{29}}=\frac{4}{29} }\\

\mathrm{\theta =\cos ^{-1}\left(\frac{4}{29}\right) \Rightarrow \theta=\sec ^{-1}\left(\frac{29}{4}\right)}

Hence the correct asnwer is option 2

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Gunjita

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