Find the angle the vector sqrt{3} hat{i} - ha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The vector is given by $\vec{A} = \sqrt{3} \hat{i} - \hat{j}$.Comparing this with $\vec{A} = A_x \hat{i} + A_y \hat{j}$,we get $A_x = \sqrt{3}$ an

Vedclass · Accepted Answer

$-30$

Vedclass · Answer

(B) The vector is given by $\vec{A} = \sqrt{3} \hat{i} - \hat{j}$.Comparing this with $\vec{A} = A_x \hat{i} + A_y \hat{j}$,we get $A_x = \sqrt{3}$ and $A_y = -1$.The angle $	heta$ that the vector makes with the positive $x$-axis is given by $	an 	heta = \frac{A_y}{A_x}$.Substituting the values,we have $	an 	heta = \frac{-1}{\sqrt{3}}$.Since $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$,$	an(-30^{\circ}) = -\frac{1}{\sqrt{3}}$.Therefore,$	heta = -30^{\circ}$.

Find the angle the vector $\sqrt{3} \hat{i} - \hat{j}$ makes with the positive $x$-axis. (in $^{\circ}$)

Similar Questions

The sum of three vectors is zero as shown in the figure. What will be the magnitudes of vectors $\vec{OB}$ and $\vec{OC}$?

$A$ displacement vector of magnitude $4$ makes an angle $30^{\circ}$ with the $x$-axis. Its rectangular components in the $x-y$ plane are .........

Add vectors $\vec{A}$,$\vec{B}$,and $\vec{C}$,each having a magnitude of $50 \text{ units}$ and inclined to the $X$-axis at angles $45^{\circ}$,$135^{\circ}$,and $315^{\circ}$ respectively.

Consider an object kept at the centre, in the $XY$-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is

The component of vector $\overrightarrow{A}=a_x \hat{i}+a_y \hat{j}+a_z \hat{k}$ along the direction of $\hat{i}-\hat{j}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module