Question

If $\overrightarrow{A}=4\hat{i}+6\hat{j}$ and $\overrightarrow{B}=2\hat{i}+3\hat{j}$ .Then :

• A. $\overrightarrow{A}.\overrightarrow{B}=29$
• B. $\overrightarrow{A}\times \overrightarrow{B}=0$
• C. $\frac{|\overrightarrow{A}|}{|\overrightarrow{B}|}=\frac{1}{1}$
• D. angles between $\overrightarrow{A}$ and $\overrightarrow{B}$ is ${30}^{\circ }$

Answer 1

The correct option is B $\overrightarrow{A}\times \overrightarrow{B}=0$

(A) $\overrightarrow{A}.\overrightarrow{B}=(4\hat{i}+6\hat{j}).(2\hat{i}+3\hat{j})=8+18=26$

(B) $\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 6& 0\\ 2& 3& 0\end{array}\right|=\hat{i}(0-0)-\hat{j}(0-0)+\hat{k}(12-12)=0$

(C) $|\overrightarrow{A}|=\sqrt{4^2+6^2}=7.21$

$|\overrightarrow{B}|=\sqrt{2^2+3^2}=3.60$

$\frac{|\overrightarrow{A}|}{|\overrightarrow{B}|}=2$

(D) Angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ $=co{s}^{-1}\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}|.|\overrightarrow{B}|}=co{s}^{-1}1=0^0$