Let $\mathbf{A}_1+\mathbf{A}_2=5 \mathbf{A}_3, \mathbf{A}_1-\mathbf{A}_2=3 \mathbf{A}_3$, $\mathbf{A}_3=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$, then $\frac{\left|\mathbf{A}_1\right|}{\left|\mathbf{A}_2\right|} \mathrm{i}$

• Correct answer
  A

  4
• B

  8
• C

  2
• D

  6

Solution

Given that,
On adding Eqs. (i) and (ii), we get $\begin{aligned} 2 \mathbf{A}_1 & =8 \mathbf{A}_3 \Rightarrow \mathbf{A}_1=4 \mathbf{A}_3 \\ \mathbf{A}_1 & =4(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})\left[\because \mathbf{A}_3=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} \text { (given) }\right] \\ \mathbf{A}_1 & =8 \hat{\mathbf{i}}+16 \hat{\mathbf{j}} \end{aligned}$ On subtracting Eqs. (ii) from (i), we get $\begin{aligned} 2 \mathbf{A}_2 & =2 \mathbf{A}_3 \Rightarrow \mathbf{A}_2=\mathbf{A}_3=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} \\ \frac{\left|\mathbf{A}_1\right|}{\left|\mathbf{A}_2\right|} & =\frac{|8 \hat{\mathbf{i}}+16 \hat{\mathbf{j}}|}{|2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}|}=\frac{\sqrt{(8)^2+(16)^2}}{(2)^2+(4)^2} \\ & =\frac{\sqrt{64+256}}{\sqrt{4+16}}=\frac{\sqrt{320}}{\sqrt{20}}=\sqrt{16}=4 \end{aligned}$

Hence, the correct answer is option: 'A'.