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 ??

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Answer 1 · 2018-03-05T16:11:42

Magnitude $94.56$

Direction $198.41^@$

2 .d.p.

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 $vecv$ :

$vecv=||v||costhetabbi+||v||sinthetabbj$

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

For $A$

$vecA=||A||cos(28)bbi+||A||sin(28)bbj$

$vecA=44cos(28)bbi+44sin(28)bbj$

For $B$

$vecB=||B||cos(124)bbi+||B||sin(124)bbj$

$vecB=26.6cos(124)bbi+26.6sin(124)bbj$

$R=B-2A$

$R=26.6cos(124)bbi+26.6sin(124)bbj-2(44cos(28)bbi+44sin(28)bbj)$

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

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

$||R||=sqrt((26.6cos(124)-88cos(28))^2+(26.6sin(124)-88sin(28))^2)$

$=94.557$

Direction:

$arctan((26.6sin(124)-88sin(28))/(26.6cos(124)-88cos(28)))=18.41^@$

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

$180^@+18.41=198.41^@$