Discussion
A straight line passes through the point whose position vector is \(\hat{k}\) . The straight line also passes through the point of intersection of the lines \(\vec{r} = \hat{j} + \lambda\hat{i}, \lambda \in \mathbb{R}\) and \(\vec{r} = \hat{i} + s\hat{j}, s \in \mathbb{R}\) . Then the equation of the straight line is
Answer: A
The line is \(\vec{r} = \hat{k} + t(\hat{i} + \hat{j} - \hat{k})\).
The two given lines are \((\lambda, 1, 0)\) and \((1, s, 0)\). They meet when \(\lambda = 1\) and \(s = 1\), at the point \((1, 1, 0)\).
The required line joins \((0, 0, 1)\) to \((1, 1, 0)\), so its direction is
\[ (1, 1, 0) - (0, 0, 1) = (1, 1, -1) \]
giving \(\vec{r} = \hat{k} + t(\hat{i} + \hat{j} - \hat{k})\). Every other option's direction, added to \(\hat{k}\), never reaches \((1, 1, 0)\); for instance \(\hat{i} - \hat{j} - \hat{k}\) gives \((1, -1, 0)\) at \(t = 1\).
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