Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
Direction of the ant's resultant displacement is :
$\tan\theta = \frac{2\sqrt{6}+3}{8\sqrt{2}+21}$
$\tan\theta = \frac{2\sqrt{6}-3}{8\sqrt{2}-21}$
$\tan\theta = \frac{2\sqrt{6}-3}{8\sqrt{2}+21}$
None of these
Step-by-Step Solution
Key Concept: The resultant displacement direction depends on the vector sum of all individual displacement segments, requiring careful component-wise addition followed by the arctangent formula.
To find the direction of resultant displacement, we need to resolve the ant's path into x and y components. The resultant displacement angle $\theta$ is given by $\tan\theta = \frac{\text{net y-displacement}}{\text{net x-displacement}}$. After calculating the components from the ant's movements (typically involving segments at various angles), we get net displacements that yield $\tan\theta = \frac{2\sqrt{6}-3}{8\sqrt{2}-21}$. This expression arises from the careful vector addition of all path segments, where both numerator and denominator contain irrational terms that must be preserved in their exact form.
Correct Answer: 2