Question

The magnitude of vectors $\vec{A},\vec{B}$ and $\vec{C}$ are $3,4$ and $5$ units respectively. If $\vec{A}+\vec{B}=\vec{C}$, the angle between $\vec{A}$ and $\vec{B}$ is

• A. $\pi /2$
• B. ${\cos }^{-1}(0.6)$
• C. ${\tan }^{-1}(7/5)$
• D. $\pi /4$

Answer 1

Answer: (A)

Let $\theta$ be angle between $\vec{A}$ and $\vec{B}$
Given: $A=|\vec{A}|=3$ units
$B=|\vec{B}|=4$ units
$C=|\vec{C}|=5$ units
$\vec{A}+\vec{B}=\vec{C}$
$\therefore (\vec{A}+\vec{B})\cdot (\vec{A}+\vec{B})=\vec{C}\cdot \vec{C}$
$\vec{A}\cdot \vec{A}+\vec{A}\cdot \vec{B}+\vec{B}\cdot \vec{A}+\vec{B}\cdot \vec{B}=\vec{C}\cdot \vec{C}$
$A^2+2AB\cos \theta +B^2=C^2$
$9+2AB\cos \theta +16=25$
or $2AB\cos \theta =0$
or $\cos \theta =0$
$\therefore \theta =90^{\circ }$.