Question

Calculate the angle between two vectors 2F and$\sqrt{2}F$ so that the resultant force is $F\sqrt{10}.$

• A. 120 degrees
• B. 90 degrees
• C. 60 degrees
• D. 45 degrees

Answer 1

The correct option is D 45 degrees

Let $\theta$ be the angle between $2F$ and $\sqrt{F}$ resultant:-

$\sqrt{(2F{)}^2+(\sqrt{2F}{)}^2+2\left(2F\right)\left(\sqrt{2}F\cos \theta \right)}=F\sqrt{10}\sqrt{4F^2+2F^2+4F^2\sqrt{2\cos \theta }}=F\sqrt{10}\sqrt{6F^2+4F^2\sqrt{2\cos \theta }}=(\sqrt{10}F{)}^{2}6F^2+4F^2\sqrt{2}\cos \theta =10F^{2}3+2\sqrt{2}\cos \theta =5\cos \theta =\frac{5-3}{2\sqrt{2}}\cos \theta =\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\theta ={\cos }^{-1}\frac{1}{\sqrt{2}}=45°$