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A line $\mathrm{L}_1$ passes through the point, whose p. v. (position vector) $3 \hat{\mathrm{i}}$, is parallel to the vector $-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$. Another line $\mathrm{L}_2$ passes through the point having p.v. $\hat{i}+\hat{j}$ is parallel to vector $\hat{i}+\hat{k}$, then the point of intersection of lines $L_1$ and $L_2$ has p.v.
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Verified Answer
The correct answer is:
$2 \hat{i}+\hat{j}+\hat{k}$
Equation of line $\mathrm{L}_1$ is $\overline{\mathrm{r}}=3 \hat{\mathrm{i}}+\lambda(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$
Equation of line $L_2$ is $\overline{r^{\prime}}=\hat{i}+\hat{j}+\lambda^{\prime}(\hat{i}+\hat{k})$
The point of intersection of $\mathrm{L}_1$ and $\mathrm{L}_2$ will satisfy $\overline{\mathrm{r}}=\overline{\mathrm{r}^{\prime}}$
$$
\begin{aligned}
& \Rightarrow 3 \hat{\mathrm{i}}+\lambda(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda^{\prime}(\hat{\mathrm{i}}+\hat{\mathrm{k}}) \\
& \Rightarrow(3-\lambda) \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}=\left(1+\lambda^{\prime}\right) \hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda^{\prime} \hat{\mathrm{k}} \\
& \Rightarrow 3-\lambda=1+\lambda^{\prime} \text { and } \lambda=1 \\
& \Rightarrow \lambda=1 \text { and } \lambda^{\prime}=1
\end{aligned}
$$
Substituting the value of $\lambda$ in (i), we get the point of intersection.
$\therefore \quad$ The point of intersection of lines $\mathrm{L}_1$ and $\mathrm{L}_2$ has p.v. $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$.
Equation of line $L_2$ is $\overline{r^{\prime}}=\hat{i}+\hat{j}+\lambda^{\prime}(\hat{i}+\hat{k})$
The point of intersection of $\mathrm{L}_1$ and $\mathrm{L}_2$ will satisfy $\overline{\mathrm{r}}=\overline{\mathrm{r}^{\prime}}$
$$
\begin{aligned}
& \Rightarrow 3 \hat{\mathrm{i}}+\lambda(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda^{\prime}(\hat{\mathrm{i}}+\hat{\mathrm{k}}) \\
& \Rightarrow(3-\lambda) \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}=\left(1+\lambda^{\prime}\right) \hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda^{\prime} \hat{\mathrm{k}} \\
& \Rightarrow 3-\lambda=1+\lambda^{\prime} \text { and } \lambda=1 \\
& \Rightarrow \lambda=1 \text { and } \lambda^{\prime}=1
\end{aligned}
$$
Substituting the value of $\lambda$ in (i), we get the point of intersection.
$\therefore \quad$ The point of intersection of lines $\mathrm{L}_1$ and $\mathrm{L}_2$ has p.v. $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$.
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