$A$ vector in the $x-y$ plane makes an angle of $30^{\circ}$ with the $y$-axis. The magnitude of the $y$-component of the vector is $2 \sqrt{3}$. The magnitude of the $x$-component of the vector will be:

  • A
    $\frac{1}{\sqrt{3}}$
  • B
    $6$
  • C
    $\sqrt{3}$
  • D
    $2$

Explore More

Similar Questions

$A$ person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?

Can the magnitude of the rectangular components of a vector be greater than the magnitude of the vector itself? Explain.

The $y$-component of vector $\vec{A}$ is $+3.0 \ m$. If $\vec{A}$ makes an angle of $30^{\circ}$ counterclockwise from the positive $y$-axis, the magnitude of $\vec{A}$ is (assume $\vec{A}$ is in the $x-y$ plane).

Three vectors $\overrightarrow{OP}, \overrightarrow{OQ}$ and $\overrightarrow{OR}$ each of magnitude $A$ are acting as shown in the figure. The resultant of the three vectors is $A \sqrt{x}$. The value of $x$ is:

$A$ force of $5\, N$ acts on a particle along a direction making an angle of $60^\circ$ with the vertical. Its vertical component is.......$N$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo