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If $\vec{r}_{1}=\lambda \hat{i}+2 \hat{j}+\hat{k}, \vec{r}_{2}=\hat{i}+(2-\lambda) \hat{j}+2 \hat{k}$ are such that
$\left|\vec{r}_{1}\right|>\left|\vec{r}_{2}\right|$, then $\lambda$ satisfies which one of the following?
Options:
$\left|\vec{r}_{1}\right|>\left|\vec{r}_{2}\right|$, then $\lambda$ satisfies which one of the following?
Solution:
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Verified Answer
The correct answer is:
$\lambda \geq 1$
Given, $\overrightarrow{\mathbf{r}}_{1}=\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\begin{aligned} & \text { and } \overrightarrow{\mathbf{r}}_{1}=\hat{\mathbf{i}}+(2-\lambda) \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \therefore\left|\overrightarrow{\mathbf{r}}_{1}\right|>\left|\overrightarrow{\mathbf{r}}_{2}\right| \\ \Rightarrow & \sqrt{\lambda^{2}+(2)^{2}+(1)^{2}}>\sqrt{(1)^{2}+(2-\lambda)^{2}+(2)^{2}} \\ \Rightarrow & \lambda^{2}+4+1>1+4+\lambda^{2}-4 \lambda+4 \\ \Rightarrow & 5>9-4 \lambda \\ \Rightarrow & 4 \lambda>4 \\ \Rightarrow & \lambda>1 \end{aligned}$
$\begin{aligned} & \text { and } \overrightarrow{\mathbf{r}}_{1}=\hat{\mathbf{i}}+(2-\lambda) \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \therefore\left|\overrightarrow{\mathbf{r}}_{1}\right|>\left|\overrightarrow{\mathbf{r}}_{2}\right| \\ \Rightarrow & \sqrt{\lambda^{2}+(2)^{2}+(1)^{2}}>\sqrt{(1)^{2}+(2-\lambda)^{2}+(2)^{2}} \\ \Rightarrow & \lambda^{2}+4+1>1+4+\lambda^{2}-4 \lambda+4 \\ \Rightarrow & 5>9-4 \lambda \\ \Rightarrow & 4 \lambda>4 \\ \Rightarrow & \lambda>1 \end{aligned}$
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