Physics - Vectors Question with Solution | TestHub

Given $\vec{A}=\vec{B}+\vec{C}$

Rearranging the terms, we get:

$\vec{B}=\vec{A}-\vec{C}$

Taking the dot product of $\vec{B}$ with itself:

$|\vec{B}{|}^2=|\vec{A}-\vec{C}{|}^2$

$|\vec{B}{|}^2=|\vec{A}{|}^2+|\vec{C}{|}^2-2|\vec{A}||\vec{C}|\cos \theta$

where $\theta$ is the angle between $\vec{A}$ and $\vec{C}$.

Substituting the given magnitudes $A=5$, $B=4$, and $C=3$:

$4^2=5^2+3^2-2(5)(3)\cos \theta$

$16=25+9-30\cos \theta$

$16=34-30\cos \theta$

$30\cos \theta =18$

$\cos \theta =\frac{18}{30}=\frac{3}{5}$

$\theta ={\cos }^{-1}\left(\frac{3}{5}\right)$

Answer: ${\cos }^{-1}\left(3/5\right)$