Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade None
Question:
Let position vector of point $A$ be $\vec{i} + \vec{j} + \vec{k}$ and that of point $B$ be $-\vec{i} + \vec{k}$, then the position vector of point $R(\vec{r})$ such that $AR$ is perpendicular to $BR$ and $\vec{r}$ is not perpendicular to $\vec{r} - (\vec{j} + 2\vec{k})$ is:
$\vec{r} = \vec{i} + 2\vec{j}$
$\vec{r} = 2\vec{i} + \vec{j} - \vec{k}$
$\vec{r} = \vec{k} + 2\vec{i}$
None of these
Step-by-Step Solution
Key Concept: Perpendicularity of two segments from fixed points creates a locus equation.
Given $\vec{AR} = \vec{r} - \vec{a}$ and $\vec{BR} = \vec{r} - \vec{b}$, the perpendicularity condition $(\vec{r} - \vec{a}) \cdot (\vec{r} - \vec{b}) = 0$ expands to $|\vec{r}|^2 - (\vec{a} + \vec{b}) \cdot \vec{r} + \vec{a} \cdot \vec{b} = 0$. This yields either $\vec{r} \perp (\vec{r} - (\vec{a} + \vec{b}))$ or $\vec{r} = \vec{j} + 2\vec{k}$ as the specific solution.
Correct Answer: 4