Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
The position vectors of the vertices $A$, $B$ and $C$ of a triangle are $\vec{i}+\vec{j}$, $\vec{j}+\vec{k}$ and $\vec{i}+\vec{k}$ respectively. A unit vector $\vec{r}$ lying in the plane of $\triangle ABC$ and perpendicular to $IA$, where $I$ is the incentre of the triangle is:
\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}
\frac{\vec{i}-\vec{j}}{\sqrt{2}}
\frac{\vec{j}-\vec{i}}{\sqrt{2}}
\frac{\vec{i}+\vec{j}-\vec{k}}{\sqrt{3}}
Step-by-Step Solution
Key Concept: For an equilateral triangle, the incentre coincides with the centroid, and the required perpendicular unit vector is found by taking the cross product of $\vec{IA}$ with the plane normal.
First, find the side lengths: $|AB| = |BC| = |CA| = \sqrt{2}$, so the triangle is equilateral. The incentre of an equilateral triangle coincides with its centroid: $I = \frac{(\vec{i}+\vec{j}) + (\vec{j}+\vec{k}) + (\vec{i}+\vec{k})}{3} = \frac{2\vec{i}+2\vec{j}+2\vec{k}}{3}$. Then $\vec{IA} = (\vec{i}+\vec{j}) - \frac{2\vec{i}+2\vec{j}+2\vec{k}}{3} = \frac{\vec{i}+\vec{j}-2\vec{k}}{3}$. The normal to the plane $ABC$ is $\vec{AB} \times \vec{AC} = (\vec{j}-\vec{i}+\vec{k}) \times (\vec{k}-\vec{i}) = \vec{i}+\vec{j}+\vec{k}$. A unit vector in the plane perpendicular to $\vec{IA}$ must be perpendicular to both $\vec{IA}$ and the plane normal. Computing $\vec{IA} \times (\vec{i}+\vec{j}+\vec{k})$ gives a direction parallel to $\vec{i}-\vec{j}$ (or $\vec{j}-\vec{i}$), which when normalized yields $\pm\frac{\vec{i}-\vec{j}}{\sqrt{2}}$.
Correct Answer: 2,3