JEE PYQ: Vector Algebra - Question ID 889117efef37 (JEE Main 2023)

When two vectors (forces, in this case) act at a point, their resultant is determined by the vector addition rule. If the two vectors are perpendicular to each other, their magnitudes and the angle between them ($90^\circ$) allow us to compute the magnitude of the resultant using the Pythagorean theorem for vectors.

The key formula is: $|\vec{R}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}$ where $\theta$ is the angle between the two vectors. When $\theta = 90^\circ$, $\cos 90^\circ = 0$, simplifying the formula to: $|\vec{R}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2}$

Given:

• Magnitude of first force: $|\vec{F_1}| = A$
• Magnitude of second force: $|\vec{F_2}| = \frac{A}{2}$
• The two forces are perpendicular, so the angle between them, $\theta = 90^\circ$.

Step 1: Write the formula for the magnitude of the resultant force when two vectors are perpendicular: $|\vec{R}| = \sqrt{|\vec{F_1}|^2 + |\vec{F_2}|^2}$

Step 2: Substitute the given magnitudes into the formula: $|\vec{R}| = \sqrt{A^2 + \left(\frac{A}{2}\right)^2}$

Step 3: Simplify the expression inside the square root: $\left(\frac{A}{2}\right)^2 = \frac{A^2}{4}$ So, $|\vec{R}| = \sqrt{A^2 + \frac{A^2}{4}} = \sqrt{\frac{4A^2}{4} + \frac{A^2}{4}} = \sqrt{\frac{5A^2}{4}}$

Step 4: Take the square root of the numerator and denominator: $|\vec{R}| = \frac{\sqrt{5A^2}}{\sqrt{4}} = \frac{\sqrt{5} \cdot A}{2}$

Step 5: Compare the derived expression with the given options. The correct answer matches option C: $\frac{\sqrt{5} A}{2}$

Trap 1: Incorrect Angle Assumption
Students often forget that the formula for the resultant magnitude depends on the angle between the vectors. If they mistakenly assume the angle is $0^\circ$ or $180^\circ$, they may use the wrong formula, leading to incorrect results like $A + \frac{A}{2} = \frac{3A}{2}$ or $A - \frac{A}{2} = \frac{A}{2}$. Always verify the angle between the vectors before applying the formula.

Trap 2: Algebraic Errors in Simplification
When simplifying $\sqrt{A^2 + \left(\frac{A}{2}\right)^2}$, students may incorrectly compute $\left(\frac{A}{2}\right)^2$ as $\frac{A}{4}$ instead of $\frac{A^2}{4}$. This leads to errors in the final expression. Always double-check the squaring of fractions.

Trap 3: Confusing Scalar and Vector Quantities
Some students may attempt to add the magnitudes directly ($A + \frac{A}{2} = \frac{3A}{2}$), ignoring the vector nature of the problem. Remember that forces are vectors, and their resultant depends on both magnitude and direction.

Exam Tip:
For problems involving perpendicular vectors, always use the Pythagorean theorem for the resultant magnitude. This is a high-yield concept in vector algebra and appears frequently in JEE exams. Practice similar problems to build intuition for vector addition.