Vector Algebra Question 132
Question: 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 $ | {{{\vec{r}}}_1} |>| {{{\vec{r}}}_2} | $ , then $ \lambda $ satisfies which one of the following?
Options:
A) $ \lambda =0 $ only
B) $ \lambda =1 $
C) $ \lambda <1 $
D) $ \lambda >1 $
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Answer:
Correct Answer: D
Solution:
- [d] Given. $ {{\vec{r}}_1}=\lambda \hat{i}+2\hat{j}+\hat{k} $ and $ {{\vec{r}}_1}=\hat{i}+(2-\lambda )\hat{j}+2\hat{k} $
$ \therefore | {{{\vec{r}}}_1} |>| {{{\vec{r}}}_2} | $
$ \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 $
BETA
