Question

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 ??

Answer 1

Magnitude `94.56`

Direction `198.41^@`

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 `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^@`