Question

magitude of position vector in 2-D is 4. its slope is 1/root 3 the position vector is

Answer 1

Explanation:

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Answer 2

The position vector is: $\vec r=2[\sqrt3\hat{i}+\hat{j}]$

Explanation:

• The Magnitude of the position vector is 4 and the slope is $\frac{1}{\sqrt3}$.
• The Slope of a vector is $tan\theta$.
• $tan\theta=\frac{1}{\sqrt3}$

           $\theta=30^\circ$

• The position is denoted as, $\vec r=rcos\theta\hat{i}+rsin\theta\hat{j}$
• $\vec r=4cos30^\circ\hat{i}+4sin30^\circ\hat{j}$
• $\vec r=4\frac{\sqrt3}{2}\hat{i}+4\frac{1}{2}\hat{j}$
• $\vec r=2\sqrt3\hat{i}+2\hat{j}\\\vec r=2[\sqrt3\hat{i}+\hat{j}]$

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