1. ## Vector Problem

An airplane flies horizontally from east to west at 304 mi/hr relative to the air. If it flies in a steady 24 mi/hr wind that blows horizontally toward the southwest (45 degrees south of west), find the speed and direction of the airplane relative to the ground.

2. ## Re: Vector Problem

Hello, Jessica!

An airplane flies horizontally from east to west at 304 mi/hr relative to the air.
If it flies in a steady 24 mi/hr wind that blows horizontally toward the southwest (45o south of west),
find the speed and direction of the airplane relative to the ground.

Code:
                         304
- - - B *   *   *   *   *   * A
45o * 135o          *
*           *
24 *       *
*   *
C*
The plane is flying from $\displaystyle A$ to $\displaystyle B$ at 304 mph.
But there is wind blowing from $\displaystyle B$ to $\displaystyle C$ at 24 mph.
Hence, the plane's flight is from $\displaystyle A$ to $\displaystyle C.$

Law of Cosines:
.$\displaystyle AC^2 \;=\;24^2 + 304^2 - 2(24)(304)\cos135^o \;=\; 103,\!210.1002$

Hence: .$\displaystyle AC \;=\;321.4188889 \;\approx\;321.4\text{ mph.}$

Law of Cosines:
. . $\displaystyle \cos A \;=\;\frac{304^2 + 321.4^2 - 24^2}{2(204)(321.4)} \;=\;0.998601718$

Hence: .$\displaystyle A \;=\;3.030300248 \;\approx\;3^o$

The direction is 3o south of west.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Um . . . it said vectors, didn't it?

$\displaystyle \overrightarrow{AB} \;=\;\langle 304\cos180^o,\:304\sin180^o\rangle \;=\;\big\langle -304,\:0\big\rangle$

$\displaystyle \overrightarrow{BC} \;=\;\langle 24\cos225^o,\:24\sin225^o\rangle \;=\;\left\langle \text{-}12\sqrt{2},\:\text{-}12\sqrt{2}\right\rangle$

$\displaystyle \overrightarrow{AC} \;=\;\overrightarrow{AB} + \overrightarrow{BC} \;=\; \big\langle \text{-}304-12\sqrt{2},\:\text{-}12\sqrt{2}\big\rangle$

Then: .$\displaystyle |\overrightarrow{AC}| \;=\;\sqrt{\left(\text{-}304 - 12\sqrt{2}\right)^2 + \left(\text{-}12\sqrt{2}\right)^2}$

Hence: .$\displaystyle |\overrightarrow{AC}| \;=\;\sqrt{103,\!310.1022} \;=\;321.418889 \;\approx\;321.4\text{ mph.}$

$\displaystyle \tan\theta \;=\;\frac{\text{-}12\sqrt{2}}{\text{-}304-12\sqrt{2}} \;=\;0.052872695$

. . $\displaystyle \theta \;=\;3.026561265 \;\approx\;3^o$

I'm gratified that my answers match!

3. ## Re: Vector Problem

Thanks, this really helped!