JEE PYQ: Vector Algebra - Question ID 42a3e5fdc289 (JEE Main 2023)

ID: 42a3e5fdc289JEE Main 2023Single Correct MCQ
A vector in xyx-y plane makes an angle of 3030^{\circ} with yy-axis. The magnitude of y\mathrm{y}-component of vector is 232 \sqrt{3}. The magnitude of xx-component of the vector will be :
JEE Question illustration 42a3e5fdc289

Select Option

Step-by-step Explanation

Core Formula & Concept:

In vector algebra, any vector A\vec{A} lying in the xx-yy plane can be resolved into its Cartesian components along the xx and yy axes. If the vector makes an angle θ\theta with the yy-axis, then:

  • The yy-component is given by Ay=AcosθA_y = |\vec{A}| \cos \theta.
  • The xx-component is given by Ax=AsinθA_x = |\vec{A}| \sin \theta.

These relations arise from the definitions of cosine and sine in a right triangle formed by the vector and its projections on the axes.

Step-by-Step Derivation:

Given:

  • Angle with yy-axis, θ=30\theta = 30^\circ.
  • Magnitude of yy-component, Ay=23A_y = 2\sqrt{3}.

Step 1: Find the magnitude of the vector A|\vec{A}|.

Using the relation for the yy-component: Ay=AcosθA_y = |\vec{A}| \cos \theta Substitute the known values: 23=Acos302\sqrt{3} = |\vec{A}| \cos 30^\circ We know that cos30=32\cos 30^\circ = \frac{\sqrt{3}}{2}, so: 23=A322\sqrt{3} = |\vec{A}| \cdot \frac{\sqrt{3}}{2} Solve for A|\vec{A}|: A=2332=2323=4|\vec{A}| = \frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2\sqrt{3} \cdot \frac{2}{\sqrt{3}} = 4

Step 2: Find the magnitude of the xx-component AxA_x.

Using the relation for the xx-component: Ax=AsinθA_x = |\vec{A}| \sin \theta Substitute the known values: Ax=4sin30A_x = 4 \cdot \sin 30^\circ We know that sin30=12\sin 30^\circ = \frac{1}{2}, so: Ax=412=2A_x = 4 \cdot \frac{1}{2} = 2

Conclusion:

The magnitude of the xx-component of the vector is 22, which corresponds to option B.

Common Traps & Exam Tip:

Students often confuse the angle given with respect to the yy-axis as the angle with the xx-axis. This leads to incorrect use of sine and cosine functions. Always:

  • Draw a clear diagram to visualize the vector and its components.
  • Identify which axis the angle is measured from (here, it is the yy-axis).
  • Use Ay=AcosθA_y = |\vec{A}| \cos \theta and Ax=AsinθA_x = |\vec{A}| \sin \theta when the angle is with the yy-axis.
  • Double-check the trigonometric values for standard angles (e.g., 3030^\circ, 4545^\circ, 6060^\circ).

A common mistake is to assume the angle is with the xx-axis and use Ax=AcosθA_x = |\vec{A}| \cos \theta, which would yield an incorrect result.