Kerala KEAM 5 June 2024 Solved Paper

We are given two parallel straight lines:1. $\vec{r} = \hat{k} + s(\hat{i} + \hat{j})$, where $s \in \mathbb{R}$2. $\vec{r} = \hat{j} + t(\hat{i} + \hat{j})$, where $t \in \mathbb{R}$We need to find the shortest distance between these two parallel lines. To solve this, we use the formula for the shortest distance $d$ between two parallel lines given by:$$d = \frac{ | \vec{r_1} - \vec{r_2} \cdot (\vec{v_1} \times \vec{v_2}) | }{ | \vec{v_1} \times \vec{v_2} | }$$Here:- $\vec{r_1} = \hat{k}$ (from the first line)- $\vec{r_2} = \hat{j}$ (from the second line)- $\vec{v_1} = \hat{i} + \hat{j}$ (direction vector of the first line)- $\vec{v_2} = \hat{i} + \hat{j}$ (direction vector of the second line)Since the lines are parallel, $\vec{v_1} = \vec{v_2}$, so the cross product $\vec{v_1} \times \vec{v_2}$ will give a zero vector. We now use the formula for the shortest distance between two parallel lines:$$d = \frac{ | (\hat{k} - \hat{j}) \cdot (\hat{i} + \hat{j}) | }{ | \hat{i} + \hat{j} | }$$We calculate $\hat{k} - \hat{j} = (0, -1, 1)$ and $\hat{i} + \hat{j} = (1, 1, 0)$, and their dot product: $$(\hat{k} - \hat{j}) \cdot (\hat{i} + \hat{j}) = (0, -1, 1) \cdot (1, 1, 0) = 0 \times 1 + (-1) \times 1 + 1 \times 0 = -1$$Now, we calculate the magnitude of $\hat{i} + \hat{j}$: $$| \hat{i} + \hat{j} | = \sqrt{1^2 + 1^2} = \sqrt{2}$$Thus, the shortest distance is: $$d = \frac{ | -1 | }{\sqrt{2}} = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad d = \frac{\sqrt{3}}{2}$$Thus, the correct answer is option (C), $\frac{\sqrt{3}}{2}$. Quick Tip: The shortest distance between two parallel lines can be calculated by finding the difference vector of a point on one line and a point on the other line, taking the dot product with the direction vector, and dividing by the magnitude of the direction vector.