What is the angle between resultant of $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$ and…
What is the angle between resultant of $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}$ ?
$\pi \mathrm{rad}$
$0^{\circ}$
$\frac{\pi}{2} \mathrm{rad}$
$\frac{\pi}{4} \mathrm{rad}$
Solution
Given,
Two vectors A and B.
A and B are two vectors. So, their sum A + B lies in the same plane where A and B lie (Since they are non -parallel so they define a plane and the cross product between them is not zero.)
$\mathrm{A} \times \mathrm{B}=|\mathrm{A}||\mathrm{B}| \sin \alpha \mathrm{n}$, where $\alpha$ is the angle between $\mathrm{A} \& \mathrm{~B}$ and $\mathrm{n}$ is the unit vector perpendicular to the plane containing A \& B. So, the angle between $(\mathrm{A}+\mathrm{B})$ and $(\mathrm{A} \times \mathrm{B})$ is $90^{\circ}$.
Mathematically,
$|\mathrm{A}+\mathrm{B} \| \mathrm{A} \times \mathrm{B}| \cos \alpha=(\mathrm{A}+\mathrm{B}) \cdot(\mathrm{A} \times \mathrm{B})$
$=A \cdot(A \times B)+B \cdot(A \times B)$
$=B \cdot(A \times A)+A \cdot(B \times B)$
$=0+0$