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JEE PYQ: Vector Algebra - Question ID ab98810cbe83 (JEE Main 2021)

ID: ab98810cbe83JEE Main 2021Single Correct MCQ

Statement I :

Two forces

\left( {\overrightarrow P + \overrightarrow Q } \right)

and

\left( {\overrightarrow P - \overrightarrow Q } \right)

where

\overrightarrow P \bot \overrightarrow Q

, when act at an angle

\theta_1

to each other, the magnitude of their resultant is

\sqrt {3({P^2} + {Q^2})}

, when they act at an angle

\theta_2

, the magnitude of their resultant becomes

\sqrt {2({P^2} + {Q^2})}

. This is possible only when

{\theta _1} < {\theta _2}

.

Statement II :

In the situation given above.

\theta_1 = 60^\circ

and

\theta_2 = 90^\circ

In the light of the above statements, choose the most appropriate answer from the options given below :-

Select Option

Step-by-step Explanation

Core Formula & Concept:

In vector algebra, when two forces $\vec{A}$ and $\vec{B}$ act at an angle $\theta$, the magnitude of their resultant $\vec{R}$ is given by the parallelogram law of vector addition: $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}.$ This formula is the cornerstone for analyzing the resultant of two vectors.

In the given problem, the two forces are $\vec{A} = \vec{P} + \vec{Q}$ and $\vec{B} = \vec{P} - \vec{Q}$, with the condition that $\vec{P} \perp \vec{Q}$. This orthogonality simplifies the magnitudes of $\vec{A}$ and $\vec{B}$: $|\vec{A}| = |\vec{P} + \vec{Q}| = \sqrt{P^2 + Q^2}, \quad |\vec{B}| = |\vec{P} - \vec{Q}| = \sqrt{P^2 + Q^2}.$ Thus, both forces have the same magnitude, $\sqrt{P^2 + Q^2}$.

Step-by-Step Derivation:

Step 1: Express the resultant magnitudes.
Using the parallelogram law for the two cases:

R_1 = \sqrt{(\sqrt{P^2 + Q^2})^2 + (\sqrt{P^2 + Q^2})^2 + 2(\sqrt{P^2 + Q^2})^2 \cos\theta_1} = \sqrt{3(P^2 + Q^2)},

R_2 = \sqrt{(\sqrt{P^2 + Q^2})^2 + (\sqrt{P^2 + Q^2})^2 + 2(\sqrt{P^2 + Q^2})^2 \cos\theta_2} = \sqrt{2(P^2 + Q^2)}.

Step 2: Simplify the equations.
Square both sides to eliminate the square roots:

2(P^2 + Q^2) + 2(P^2 + Q^2)\cos\theta_1 = 3(P^2 + Q^2),

2(P^2 + Q^2) + 2(P^2 + Q^2)\cos\theta_2 = 2(P^2 + Q^2).

Step 3: Solve for $\cos\theta_1$ and $\cos\theta_2$.
Divide both equations by $2(P^2 + Q^2)$:

1 + \cos\theta_1 = \frac{3}{2} \implies \cos\theta_1 = \frac{1}{2},

1 + \cos\theta_2 = 1 \implies \cos\theta_2 = 0.

Step 4: Determine $\theta_1$ and $\theta_2$.
From the cosine values:

\theta_1 = 60^\circ, \quad \theta_2 = 90^\circ.

Step 5: Verify Statement I.
Since $\cos\theta_1 = \frac{1}{2}$ and $\cos\theta_2 = 0$, it follows that $\theta_1 = 60^\circ < 90^\circ = \theta_2$. Thus, Statement I is true.

Step 6: Verify Statement II.
The derived values $\theta_1 = 60^\circ$ and $\theta_2 = 90^\circ$ match exactly with Statement II. Thus, Statement II is also true.

Common Traps & Exam Tip:

Students often make the following mistakes:

Ignoring orthogonality: Forgetting that $\vec{P} \perp \vec{Q}$ leads to incorrect magnitudes for $\vec{P} + \vec{Q}$ and $\vec{P} - \vec{Q}$. Always verify the magnitudes using the Pythagorean theorem when vectors are perpendicular.
Misapplying the parallelogram law: Some students confuse the formula for the resultant, especially the sign of the cosine term. Remember that the angle $\theta$ is the angle between the two vectors, not necessarily the angle with respect to a reference axis.
Algebraic errors: Squaring the resultant expressions and simplifying can lead to sign errors or incorrect cancellations. Double-check each algebraic step to avoid such pitfalls.
Angle comparison: Students may incorrectly assume that a larger resultant implies a larger angle, which is not always true. Always solve for the angles explicitly using the cosine values.

Exam Tip: When dealing with vector resultants, always start by writing down the magnitudes of the individual vectors and then apply the parallelogram law. This systematic approach minimizes errors and ensures clarity.

Conclusion:

Both Statement I and Statement II are true, making option B the correct answer.

JEE PYQ: Vector Algebra - Question ID ab98810cbe83 (JEE Main 2021)

Select Option

Step-by-step Explanation

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