# A vector has component A_x=-3.50m and A_y=6.00m. Find the magnitude (in m) and the direction (in...

## Question:

A vector has component A{eq}_x {/eq}=-3.50m and A{eq}_y {/eq}=6.00m. Find the magnitude (in m) and the direction (in degrees counterclockwise from the +x-axis) of the vector.

## Vector Quantity:

The physical quantities which have both magnitude and direction and obey laws of vector addition are called vector quantities.

Example: displacement, acceleration, force etc.

Given Data:

• Component of vector {eq}A {/eq} in x direction, {eq}{A_x} = - 3.5\;{\rm{m}} {/eq}
• Component of vector {eq}A {/eq} in y direction, {eq}{A_y} = 6\;{\rm{m}} {/eq}

The magnitude of the vector {eq}A {/eq} is,

{eq}v = \sqrt {v_x^2 + v_y^2} {/eq}

Substitute the values in the above equation.

{eq}\begin{align*} v &= \sqrt {{{\left( { - 3.5} \right)}^2} + {{\left( 6 \right)}^2}} \\ &= \sqrt {48.25} \\ &= 6.94\;{\rm{m}} \end{align*} {/eq}

Thus, the magnitude of vector is {eq}6.94\;{{\rm{m}}^{\rm{2}}} {/eq}.

The direction of the vector {eq}A {/eq},

{eq}\theta = {\tan ^{ - 1}}\left( {\dfrac{{{A_y}}}{{{A_x}}}} \right) {/eq}

Substitute the values in the above equation.

{eq}\begin{align*} \theta &= {\tan ^{ - 1}}\left( {\dfrac{6}{{ - 3.5}}} \right)\\ &= - 59.743^\circ \end{align*} {/eq}

The angle, {eq}- 59.743^\circ {/eq} is from - x axis to + y axis in clockwise direction.

The direction of the vector from +x-axis counter-clockwise direction,

{eq}\begin{align*} \alpha &= 180^\circ - \theta \\ \alpha &= 180^\circ - 59.743^\circ \\ \alpha &= 120.257^\circ \end{align*} {/eq}

Thus, the direction of the vector from +x-axis counter-clockwise direction is {eq}120.257^\circ {/eq}. 