MHT CET · Maths · Three Dimensional Geometry
The cosine of the angle included between the lines \(\overline{\mathrm{r}}=(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}-2 \hat{k})\) and \(\overline{\mathrm{r}}=(\hat{\imath}+\hat{\jmath}+3 \hat{k})+\mu(3 \hat{\imath}+2 \hat{\jmath}-6 \hat{k})\)
where \(\lambda, \mu \in \mathrm{R}\) is
- A \(\frac{13}{21}\)
- B \(\frac{11}{21}\)
- C \(\frac{3}{21}\)
- D \(\frac{17}{21}\)
Answer & Solution
Correct Answer
(B) \(\frac{11}{21}\)
Step-by-step Solution
Detailed explanation
(D)
\(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3} \cdot \frac{1}{5}}\right)=\tan ^{-1}\) \(\left(\frac{\frac{8}{15}}{\frac{15-1}{15}}\right)=\tan ^{-1}\left(\frac{8}{14}\right)=\tan ^{-1}\left(\frac{4}{7}\right)\) \(\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} \times \frac{1}{8}}\right)=\tan ^{-1}\) \(\left(\frac{\frac{15}{56}}{\frac{55}{56}}\right)=\tan ^{-1}\left(\frac{15}{55}\right)=\tan ^{-1}\left(\frac{3}{11}\right)\) Now,
\(\tan ^{-1}\left(\frac{4}{7}\right)+\tan ^{-1}\left(\frac{3}{11}\right) =\tan ^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7} \times \frac{3}{11}}\right)=\tan ^{-1}\) \(\left(\frac{\frac{44+21}{77}}{\frac{77-12}{77}}\right)=\tan ^{-1}\left(\frac{\frac{65}{77}}{\frac{65}{77}}\right) \)
\( =\tan ^{-1}(1)=\frac{\pi}{4}\)
\(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3} \cdot \frac{1}{5}}\right)=\tan ^{-1}\) \(\left(\frac{\frac{8}{15}}{\frac{15-1}{15}}\right)=\tan ^{-1}\left(\frac{8}{14}\right)=\tan ^{-1}\left(\frac{4}{7}\right)\) \(\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} \times \frac{1}{8}}\right)=\tan ^{-1}\) \(\left(\frac{\frac{15}{56}}{\frac{55}{56}}\right)=\tan ^{-1}\left(\frac{15}{55}\right)=\tan ^{-1}\left(\frac{3}{11}\right)\) Now,
\(\tan ^{-1}\left(\frac{4}{7}\right)+\tan ^{-1}\left(\frac{3}{11}\right) =\tan ^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7} \times \frac{3}{11}}\right)=\tan ^{-1}\) \(\left(\frac{\frac{44+21}{77}}{\frac{77-12}{77}}\right)=\tan ^{-1}\left(\frac{\frac{65}{77}}{\frac{65}{77}}\right) \)
\( =\tan ^{-1}(1)=\frac{\pi}{4}\)
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