Let \(\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\quad \overrightarrow{\mathrm{b}}=2…

Let \(\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\quad \overrightarrow{\mathrm{b}}=2 \hat{i}+7 \hat{j}+3 \hat{k} . \quad\) Let \(\mathrm{L}_1: \overrightarrow{\mathrm{r}}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda \overrightarrow{\mathrm{a}}, \lambda \in \mathbf{R}\) and \(\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(\hat{j}+\hat{k})+\mu \overrightarrow{\mathrm{b}}, \mu \in \mathbf{R}\) be two lines. If the line \(\mathrm{L}_3\) passes through the point of intersection of \(\mathrm{L}_1\) and \(L_2\), and is parallel to \(\vec{a}+\vec{b}\), then \(L_3\) passes through the point :
  1. \((5,17,4)\)
  2. \((2,8,5)\)
  3. \((8,26,12)\)
  4. \((-1,-1,1)\)

Solution

$\begin{aligned} & L_1: \overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{r}}=(\lambda-1) \hat{\mathrm{i}}+2(\lambda+1) \hat{\mathrm{j}}+(\lambda+1) \hat{\mathrm{k}} \\ & \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{r}}=2 \mu \hat{\mathrm{i}}+(1+7 \mu) \hat{\mathrm{j}}+(1+3 \mu) \hat{\mathrm{k}}\end{aligned}$
For point of intersection equating respective components
$\begin{aligned}
& \Rightarrow \lambda-1=2 \mu \\ & 2(\lambda+1)=1+7 \mu \\ & \lambda+1=1+3 \mu
\end{aligned}$
We get
$\begin{aligned}
& \Rightarrow \lambda=3 \text { and } \mu=1 \\ & \Rightarrow \vec{a}+\vec{b}=3 \hat{i}+9 \hat{j}+4 \hat{k}
\end{aligned}$
$\mathrm{L}_3: \overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}+\alpha(3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$
For $\alpha=2, \overrightarrow{\mathrm{r}}=8 \hat{\mathrm{i}}+26 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}$ ^

Asked in: JEE Main 2025 (29 Jan Shift 1)

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