Question

Let $\vec{OP}=2\hat{j}$ be the position vector of 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:

• A. $\frac{\sqrt{2}}{2}$
• B. $\frac{\sqrt{3}}{3}$
• C. $\frac{\sqrt{6}}{3}$
• D. $\frac{\sqrt{2}}{3}$
• E. $\frac{\sqrt{2}}{4}$

Hint:

In 2D, the distance can also be simplified by finding the projection of the vector onto the line's normal.

Answer 1

The Correct Option is A

Step 1: Concept
Distance from point $\vec{a}$ to line $\vec{r} = \vec{b} + \lambda \vec{v}$ is $\frac{|(\vec{a}-\vec{b}) \times \vec{v}|}{|\vec{v}|}$.

Step 2: Analysis
$\vec{a} = 2\hat{j}$, $\vec{b} = \hat{j}$, $\vec{v} = \hat{i} + \hat{j}$. $\vec{a} - \vec{b} = \hat{j}$. $(\vec{a} - \vec{b}) \times \vec{v} = \hat{j} \times (\hat{i} + \hat{j}) = -\hat{k}$.

Step 3: Calculation
Distance $= \frac{|-\hat{k}|}{|\hat{i} + \hat{j}|} = \frac{1}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. Final Answer: (A)