Question

The magnitude of a position vector in a XY plane is 4. Its slope is $\dfrac{1}{{\sqrt 3 }}$, then the position vector is:
(A) $\sqrt 3 \hat i + \hat j$
(B) $2\sqrt 3 \hat i - 2\hat j$
(C) $2\sqrt 3 \hat i + 2\hat j$
(D) $2\hat i + 2\sqrt 3 \hat j$

Answer 1

Hint
The slope of a position vector is given by the tan of the angle that the vector makes with the X-axis. So from the given slope, we can find the angle. The x-component of the vector is given by the product of the magnitude and cosine of the angle and the y component is the product of magnitude and the sine of the angle.

Formula Used: In this solution, we will be using the following formula,
$m = \tan \theta$
where $m$ is the slope of the vector and $\theta$ is the angle that the vector makes with the X-axis.

Complete step by step answer
Here we are provided the slope of the vector and the magnitude of the vector on the XY plane.
Let us consider a vector $\vec A$ on the XY plane making an angle $\theta$ with the X-axis. So the slope of the vector is given by,
$m = \tan \theta$
In the question, we are said that the slope is $\dfrac{1}{{\sqrt 3 }}$. So substituting that in the value we get,
$\dfrac{1}{{\sqrt 3 }} = \tan \theta$
So from here, we can find $\theta$ by taking ${\tan ^{ - 1}}$ on both sides.
Hence we get,
${\tan ^{ - 1}}\dfrac{1}{{\sqrt 3 }} = \theta$
Therefore we get the angle as $\theta = 30^\circ$.
Since the vector lies on the XY plane so it has 2 components and can be written in vector form as,
$\vec A = x\hat i + y\hat j$ where the $\hat i$ and $\hat j$ are the unit vectors along the X and Y-axis.
Therefore the X component is given by the product of the magnitude of the vector and the cosine of the angle made on the X-axis.
$\therefore x = \left| {\vec A} \right|\cos \theta$
We are given $\left| {\vec A} \right| = 4$.
So, $x = 4 \times \cos 30^\circ$
The value of $\cos 30^\circ$ is $\dfrac{{\sqrt 3 }}{2}$.
therefore, the X-component is given by,
$x = 4 \times \dfrac{{\sqrt 3 }}{2} = 2\sqrt 3$
And the Y component is given by the product of the magnitude of the vector and the sine of the angle made on the X-axis.
$\therefore y = \left| {\vec A} \right|\sin \theta$
Substituting the values we get,
$y = 4 \times \sin 30^\circ$
The value of $\sin 30^\circ$ is $\dfrac{1}{2}$.
On substituting the value,
$y = 4 \times \dfrac{1}{2} = 2$
So substituting the values of $x$ and $y$ we get the vector as,
$\vec A = 2\sqrt 3 \hat i + 2\hat j$
Therefore the correct answer is option C.

Note
The magnitude of the X component of the vector is given by the product of the magnitude of the vector and the cosine of the angle made on the X-axis because, if we do a dot product of the vector with the unit vector along the X-axis, we get,
$\vec A \cdot \hat i = \left( {x\hat i + y\hat j} \right) \cdot \hat i$
On doing the dot product on both the sides, we get,
$\left| {\vec A} \right|\left| {\hat i} \right|\cos \theta = x\hat i \cdot \hat i$
Since the magnitude of the unit vector is 1,
$x = \left| {\vec A} \right|\cos \theta$.