Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12

Question:

Magnitude of the resultant displacement is given by :
$\sqrt{301+186\sqrt{2}}$
$\sqrt{301-168\sqrt{2}-6\sqrt{6}}$
$\sqrt{301+6\sqrt{6}}$
None of these

Step-by-Step Solution

Key Concept: The resultant magnitude depends on both the individual vector magnitudes AND the angles between them, requiring careful calculation of cross terms in the expansion of $(\vec{A} + \vec{B} + ...)^2$.
To find the resultant displacement, we typically use vector addition. When multiple displacement vectors are given at various angles, we resolve them into components and use the formula $R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$. The expression $\sqrt{301-168\sqrt{2}-6\sqrt{6}}$ arises from carefully computing the dot products of vectors at specific angles (likely involving $45°$ and $30°$ angles based on the $\sqrt{2}$ and $\sqrt{6}$ terms). The negative signs indicate that certain displacement components oppose each other, reducing the net magnitude.
Correct Answer: 2

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