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The equation of the line passing through \((0, 0, 1)\) and (1,1,0) is

  • A.\(\vec{r} = \hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k}),\) \(\lambda \in \mathbb{R}\)
  • B.\(\vec{r} = \hat{j} + \lambda(\hat{i} - \hat{j} + \hat{k}),\) \(\lambda \in \mathbb{R}\)
  • C.\(\vec{r} = \hat{i} + \lambda(\hat{i} + \hat{j} + \hat{k}),\) \(\lambda \in \mathbb{R}\)
  • D.\(\vec{r} = \hat{i} + \hat{j} + \lambda(\hat{i} - \hat{j} - \hat{k}),\) \(\lambda \in \mathbb{R}\)
  • E.\(\vec{r} = \hat{i} + \hat{j} + \hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k}),\) \(\lambda \in \mathbb{R}\)

Answer: A

The line is \(\vec{r} = \hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})\).

Take the point (0, 0, 1), with position vector \(\hat{k}\), and the direction from it to (1, 1, 0):

\[ \begin{aligned} (1, 1, 0) - (0, 0, 1) &= (1, 1, -1) \\ &= \hat{i} + \hat{j} - \hat{k} \end{aligned} \]

Check: \(\lambda = 1\) gives (1, 1, 0). The other options either start from a point not on the line (\(\hat{j}\), \(\hat{i}\), \(\hat{i} + \hat{j} + \hat{k}\)) or use a direction not parallel to (1, 1, −1).

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