Question

The resultant of the two vectors having magnitude $2$ and $3$ is $1$. What is their cross product?

• A. $6$
• B. $3$
• C. $1$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for the correct option:

Find the cross product

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be the two vectors such that,

$\left|\overrightarrow{a}\right|=2$ and $\left|\overrightarrow{b}\right|=3$

Given that the resultant is $1$

$$\begin{gathered} \therefore \sqrt{{\left(\left|\overrightarrow{a}\right|\right)}^2+{\left(\left|\overrightarrow{b}\right|\right)}^2+2·\left(\left|\overrightarrow{a}\right|\right)\left(\left|\overrightarrow{b}\right|\right)\cos \left(\theta \right)}=1 \\ \Rightarrow \sqrt{{\left(2\right)}^2+{\left(3\right)}^2+2·\left(2\right)\left(3\right)\cos \left(\theta \right)}=1 \\ \Rightarrow 13+12\cos \left(\theta \right)=1 \\ \Rightarrow \cos \left(\theta \right)=-\frac{12}{12} \\ \Rightarrow \cos \left(\theta \right)=-1 \\ \Rightarrow \theta =\pi \left[\because \cos \left(\pi \right)=-1\right] \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \left(\theta \right) \\ =2·3·\sin \left(\pi \right) \\ =0\left[\because \sin \left(\pi \right)=0\right] \end{gathered}$$

Hence, the correct option is $\left(\mathrm{D}\right)$.