MHT CET · Maths · Vector Algebra
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
- A \((-3,1,-1)\)
- B \((-3,-1,1)\)
- C \((3,1,-1)\)
- D \((3,1,1)\)
Answer & Solution
Correct Answer
(C) \((3,1,-1)\)
Step-by-step Solution
Detailed explanation
\(\vec{r} \times \vec{a}=\vec{b} \times \vec{a}\) 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 \(\lambda=1, \vec{r}=3 \hat{i}+\hat{j}-\widehat{k} \equiv(3,1,-1)\)
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