Question

$\overrightarrow{OA}=3i+4j+12k,\overrightarrow{OB}=12i+3j+4k,$ $\overrightarrow{OC}=4i+12j+3k.$
Find the angle between $\overrightarrow{OA}$ and $\overrightarrow{OC}.$

• A. $\theta ={\cos }^{-1}\frac{96}{169}$
• B. $\theta ={\sin }^{-1}\frac{96}{169}$
• C. $\theta ={\cos }^{-1}\frac{139}{169}$
• D. $\theta ={\sin }^{-1}\frac{139}{169}$

Answer 1

The correct option is A $\theta ={\cos }^{-1}\frac{96}{169}$

$\overrightarrow{OA}.\ddot{\overrightarrow{OC}}=(3i+4j+12k).(4i+12j+3k)$ .

$\therefore \left|\overrightarrow{OA}\right|$ $\left|\overrightarrow{OC}\right|\cos \theta =3.41+4.12+12.3=96$

or $\cos \theta =\frac{96}{\sqrt{(3^2+4^2+{12}^2)}.\sqrt{(4^2+{12}^2+3^2)}}=\frac{96}{169}$

$\therefore \theta ={\cos }^{-1}\frac{96}{169}.$