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If $\vec{a}$ is a vector such that $\vec{a} \times \hat{i}=\hat{j}+\hat{k}$ and $\vec{a} \cdot \hat{i}=1$, then equation of the line passing through the point $\hat{i}+\hat{j}+\hat{k}$ and parallel to $\vec{a}$ is
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The correct answer is:
$\vec{r}=(t+1) \hat{i}+(1-t) \hat{j}+(t+1) \hat{k}$
Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$
$\begin{array}{ll}
\because & \vec{a} \times \hat{i}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
x & y & z \\
1 & 0 & 0
\end{array}\right|=z \hat{j}-y \hat{k} \\
\because \quad & z \hat{j}-y \hat{k}=\hat{j}+\hat{k} \Rightarrow z=1, y=-1 \Rightarrow \vec{a} \cdot \hat{i}=1 \\
& (x \hat{i}+y \hat{j}+z \hat{k}) \cdot \hat{i}=1 \Rightarrow x=1 \\
\therefore \quad & \vec{a}=\hat{i}-\hat{j}+\hat{k} .
\end{array}$
$\therefore \quad$ Equation of line passing through $\vec{p}$ and parallel to $\vec{a}$ is
$\begin{aligned}
\vec{r} & =\vec{p}+\lambda \vec{a} \\
\Rightarrow \quad \vec{r} & =\hat{i}+\hat{j}+\hat{k}+\lambda(\hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =(t+1) \hat{i}+(1-t) \hat{j}+(t+1) \hat{k}
\end{aligned}$
$\begin{array}{ll}
\because & \vec{a} \times \hat{i}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
x & y & z \\
1 & 0 & 0
\end{array}\right|=z \hat{j}-y \hat{k} \\
\because \quad & z \hat{j}-y \hat{k}=\hat{j}+\hat{k} \Rightarrow z=1, y=-1 \Rightarrow \vec{a} \cdot \hat{i}=1 \\
& (x \hat{i}+y \hat{j}+z \hat{k}) \cdot \hat{i}=1 \Rightarrow x=1 \\
\therefore \quad & \vec{a}=\hat{i}-\hat{j}+\hat{k} .
\end{array}$
$\therefore \quad$ Equation of line passing through $\vec{p}$ and parallel to $\vec{a}$ is
$\begin{aligned}
\vec{r} & =\vec{p}+\lambda \vec{a} \\
\Rightarrow \quad \vec{r} & =\hat{i}+\hat{j}+\hat{k}+\lambda(\hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =(t+1) \hat{i}+(1-t) \hat{j}+(t+1) \hat{k}
\end{aligned}$
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