# Vector A, |A| = 44m @ 28° to + x-axis, and Vector B, |B| = 26.6m @ 124° to the + x-axis. Determine the magnitude and direction of R = B - 2A ??

Mar 5, 2018

Magnitude $94.56$

Direction ${198.41}^{\circ}$

2 .d.p.

#### Explanation:

I will do this problem using unit vectors:

We know the magnitude of $A$ and $B$.

Given that we can look at the magnitude of the vector as being the hypotenuse of a right triangle we know that for a vector $\vec{v}$:

$\vec{v} = | | v | | \cos \theta \boldsymbol{i} + | | v | | \sin \theta \boldsymbol{j}$

We need to pay attention to which quadrant we are in, so we know were values will be positive or negative:

For $A$

$\vec{A} = | | A | | \cos \left(28\right) \boldsymbol{i} + | | A | | \sin \left(28\right) \boldsymbol{j}$

$\vec{A} = 44 \cos \left(28\right) \boldsymbol{i} + 44 \sin \left(28\right) \boldsymbol{j}$

For $B$

$\vec{B} = | | B | | \cos \left(124\right) \boldsymbol{i} + | | B | | \sin \left(124\right) \boldsymbol{j}$

$\vec{B} = 26.6 \cos \left(124\right) \boldsymbol{i} + 26.6 \sin \left(124\right) \boldsymbol{j}$

$R = B - 2 A$

$R = 26.6 \cos \left(124\right) \boldsymbol{i} + 26.6 \sin \left(124\right) \boldsymbol{j} - 2 \left(44 \cos \left(28\right) \boldsymbol{i} + 44 \sin \left(28\right) \boldsymbol{j}\right)$

R=26.6cos(124)bbi+26.6sin(124)bbj-88cos(28)bbi-88sin(28)bbj)

$R = \left(26.6 \cos \left(124\right) - 88 \cos \left(28\right)\right) \boldsymbol{i} + \left(26.6 \sin \left(124\right) - 88 \sin \left(28\right)\right) \boldsymbol{j}$

$| | R | | = \sqrt{{\left(26.6 \cos \left(124\right) - 88 \cos \left(28\right)\right)}^{2} + {\left(26.6 \sin \left(124\right) - 88 \sin \left(28\right)\right)}^{2}}$

$= 94.557$

Direction:

$\arctan \left(\frac{26.6 \sin \left(124\right) - 88 \sin \left(28\right)}{26.6 \cos \left(124\right) - 88 \cos \left(28\right)}\right) = {18.41}^{\circ}$

Because the signs of the components of $R$ are negative we are in the III quadrant, so:

${180}^{\circ} + 18.41 = {198.41}^{\circ}$