Question 113: Motion in 2D & Vectors

Question

What is the angle between resultant of $A+B$ and $\mathbf{A} 	imes \mathbf{B}$.

Accepted Answer

The vector $(A+B)$ will be in plane containing vector $A$ and vector $B$.
Vector $(\mathbf{A} 	imes \mathbf{B})$ will be perpendicular to the plane containing vector $A$ and $B$.
Thus, the angle between $(A+B)$ and $(A 	imes B)$ is $90^{\circ}$ or $\frac{\pi}{2} \mathrm{rad}$.
Alternate solution
$$\begin{aligned}
&(A+B) \cdot(A 	imes B)=|A+B||A 	imes B| \cos \alpha \
& A \cdot(A 	imes B)+B \cdot(A 	imes B)=|A+B||A 	imes B| \cos \alpha \
& 0+0=|A+B||A 	imes B| \cos \alpha \
& \Rightarrow \quad \cos \alpha=0 \
& \alpha=90^{\circ} 	ext { or } \frac{\pi}{2} \mathrm{rad}
\end{aligned}$$

Answer

$\pi \mathrm{rad}$

Answer

$0^{\circ}$

Answer

$\frac{\pi}{4} \mathrm{rad}$

Answer

$\frac{\pi}{2} \mathrm{rad}$

Motion in 2D & Vectors

What is the angle between resultant of A+B and \mathbf{A} \times \mathbf{B}.

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