Question

If $\overrightarrow{V}$ is a $3-$dimensional vector satisfying $2\overrightarrow{V}+\overrightarrow{V}\times (\hat{i}+2\hat{j})=2\hat{i}+\hat{k},$ then the value of $9|\overrightarrow{V}{|}^2$ is

Answer 1

Given : $2\overrightarrow{V}+\overrightarrow{V}\times (\hat{i}+2\hat{j})=2\hat{i}+\hat{k}\cdots \left(1\right)$

Taking dot product with $\hat{i}+2\hat{j}$ on both sides, we get
$2\overrightarrow{V}\cdot (\hat{i}+2\hat{j})+\overrightarrow{0}=(2\hat{i}+\hat{k})\cdot (\hat{i}+2\hat{j})\Rightarrow 2\overrightarrow{V}\cdot (\hat{i}+2\hat{j})=2\Rightarrow \overrightarrow{V}\cdot (\hat{i}+2\hat{j})=1\Rightarrow |\overrightarrow{V}|\left|\right(\hat{i}+2\hat{j}\left)\right|\cos \theta =1\Rightarrow |\overrightarrow{V}|=\frac{1}{\sqrt{5}\cos \theta }\cdots \left(2\right)$

Now, using equation $\left(1\right)$, we get
$|2\overrightarrow{V}+\overrightarrow{V}\times (\hat{i}+2\hat{j}){|}^2=|2\hat{i}+\hat{k}{|}^2\Rightarrow 4|\overrightarrow{V}{|}^2+|\overrightarrow{V}\times (\hat{i}+2\hat{j}){|}^2+4\overrightarrow{V}\cdot (\overrightarrow{V}\times (\hat{i}+2\hat{j}\left)\right)=5\Rightarrow 4|\overrightarrow{V}{|}^2+|\overrightarrow{V}{|}^2{|\hat{i}+2\hat{j}|}^2{\sin }^2\theta =5\Rightarrow |\overrightarrow{V}{|}^2[4+5{\sin }^2\theta ]=5\Rightarrow \frac{4+5(1-{\cos }^2\theta )}{5{\cos }^2\theta }=5\Rightarrow {\cos }^2\theta =\frac{9}{30}$

From equation $\left(2\right)$, we get
$|\overrightarrow{V}{|}^2=\frac{1}{5{\cos }^2\theta }\Rightarrow |\overrightarrow{V}{|}^2=\frac{6}{9}\therefore 9|\overrightarrow{V}{|}^2=6$