Discussion
Let \(\vec{OP} = 2\hat{j}\) be the position vector a point P. Let \(\vec{r} = \hat{j} + \lambda(\hat{i} + \hat{j})\) be a straight line. The distance of the point P from the line is
Answer: A
The distance is \(\frac{\sqrt{2}}{2}\).
The line passes through A with position vector \(\hat{j}\) in the direction \(\vec{d} = \hat{i} + \hat{j}\). The distance of P from it is \(\frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}\).
\[ \begin{aligned} \vec{AP} &= 2\hat{j} - \hat{j} = \hat{j} \\ \hat{j} \times (\hat{i} + \hat{j}) &= -\hat{k} \\ d &= \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \end{aligned} \]
A quick check in the plane: the line is y = x + 1, and the distance of (0, 2) from x - y + 1 = 0 is \(\frac{|0 - 2 + 1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}\). The other values come from dividing by \(\sqrt{3}\) or 3 instead of \(\sqrt{2}\).
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