Search any question & find its solution
Question:
Answered & Verified by Expert
If $\bar{a}=\hat{i}+\hat{j}, \bar{b}=2 \hat{i}-\hat{k}$, then point of intersection of the lines $\bar{r} \times \bar{a}=\bar{b} \times \bar{a}$ and $\bar{r} \times \bar{b}=\bar{a} \times \bar{b}$ is
Options:
Solution:
2734 Upvotes
Verified Answer
The correct answer is:
$(3,1,-1)$
$\vec{r} \times \vec{a}=\vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b}=\vec{a} \times \vec{b}$
$\begin{aligned} & \Rightarrow \vec{r} \times \vec{a}=-\vec{r} \times \vec{b} \quad \text { [from (i) and (ii)] } \\ & \Rightarrow \vec{r} \times(\vec{a}+\vec{b})=\overrightarrow{0} \\ & \Rightarrow \vec{r} \| \vec{a}+\vec{b} \\ & \Rightarrow \vec{r}=\lambda(\vec{a}+\vec{b})=\lambda(3 \hat{i}+\hat{j}-\hat{k})\end{aligned}$
Taking $\lambda=1, \vec{r}=3 \hat{i}+\hat{j}-\widehat{k} \equiv(3,1,-1)$
$\begin{aligned} & \Rightarrow \vec{r} \times \vec{a}=-\vec{r} \times \vec{b} \quad \text { [from (i) and (ii)] } \\ & \Rightarrow \vec{r} \times(\vec{a}+\vec{b})=\overrightarrow{0} \\ & \Rightarrow \vec{r} \| \vec{a}+\vec{b} \\ & \Rightarrow \vec{r}=\lambda(\vec{a}+\vec{b})=\lambda(3 \hat{i}+\hat{j}-\hat{k})\end{aligned}$
Taking $\lambda=1, \vec{r}=3 \hat{i}+\hat{j}-\widehat{k} \equiv(3,1,-1)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.