Question

$\mathrm{If}\overrightarrow{A}\mathrm{and}\overrightarrow{B}\mathrm{are}\mathrm{two}\mathrm{unit}\mathrm{vectors}\mathrm{such}\mathrm{that}3\overrightarrow{A}+4\overrightarrow{B}\mathrm{and}5\overrightarrow{A}-3\overrightarrow{B}\mathrm{are}\mathrm{perpendicular},\mathrm{then}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{A}\mathrm{and}\overrightarrow{B}\mathrm{is}$

• A. ${90}^o$
• B. ${\cos }^{-1}\left(\frac{-3}{11}\right)$
• C. ${\sin }^{-1}\left(\frac{-3}{11}\right)$
• D. $0^o$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{-3}{11}\right)$

$\mathrm{Since},3\overrightarrow{A}+4\overrightarrow{B}\mathrm{and}5\overrightarrow{A}-3\overrightarrow{B}\mathrm{are}\mathrm{perpendicular}.\mathrm{So}\left(3\overrightarrow{A}+4\overrightarrow{B}\right).\left(5\overrightarrow{A}-3\overrightarrow{B}\right)=0\Rightarrow 15{\left(\overrightarrow{A}\right|}^2-9\overrightarrow{A}.\overrightarrow{B}+20\overrightarrow{A}.\overrightarrow{B}-12{\left(\overrightarrow{B}\right|}^2=0\left(\overrightarrow{A}\mathrm{and}\overrightarrow{B}\mathrm{are}\mathrm{unit}\mathrm{vectors}\right)\Rightarrow 15+11\overrightarrow{A}\overrightarrow{.B}-12=0\Rightarrow \overrightarrow{A}.\overrightarrow{B}=\frac{-3}{11}\Rightarrow \left(\overrightarrow{A}\right|\left(\overrightarrow{B}\right|\cos \theta =\frac{-3}{11}\left(\overrightarrow{A}\mathrm{and}\overrightarrow{B}\mathrm{are}\mathrm{unit}\mathrm{vectors}\right)\Rightarrow \cos \theta =\frac{-3}{11}\Rightarrow \theta ={\cos }^{-1}\left(\frac{-3}{11}\right)$