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
$\vec{A}+\vec{B}=\vec{C}$
$\therefore (\vec{A}+\vec{B}) \cdot(\vec{A}+\vec{B})=\vec{C} \cdot \vec{C}$
$\vec{A} \cdot \vec{A}+\vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{A}+\vec{B} \cdot \vec{B}=\vec{C} \cdot \vec{C}$
$A^{2}+2 A B \cos \theta+B^{2}=C^{2}$
$9+2 A B \cos \theta+16=25$
or $2 A B \cos \theta=0$
or $\cos \theta=0$
$ \therefore \theta=90^{\circ}$.