# Thread: Exam Review - Geometric Vectors

1. ## Exam Review - Geometric Vectors

A light plane is travelling at 175km/h on a heading of N8°E in a 40km/h wind from N80°E. Determine the plane's ground velocity using geometric vectors.

$\displaystyle |r| = \sqrt {175^2 + 40^2 - 2(175)(40)cos144}$

$\displaystyle |r| = 208.69km/h$

$\displaystyle \frac {sin144}{208.69} =$ $\displaystyle \frac {sinx}{40}$

$\displaystyle x = 6.47°$

Therefore, the plane's ground velocity is 208.69km/h on a bearing of N6.47°E.

2. Hello, Macleef!

A light plane is travelling at 175 km/h on a heading of N8°E
in a 40 km/h wind from N80°E.
Determine the plane's ground velocity using geometric vectors.

$\displaystyle |r| \:=\: \sqrt {175^2 + 40^2 - 2(175)(40)\cos{\color{red}144}}$ . ??
Code:
o B
* *
40   * 72°*
*       *
C    *          *
o             *
*     :     * 175
*    :    *
r  *   :   *
*  :8°*
* : *
*:*
o
A

We find that: .$\displaystyle \angle B \:=\:72^o$

Then: .$\displaystyle r^2 \:=\:40^2 + 175^2 - 2(40)(175)\cos72^o \:=\:27,898.76208$

Therefore: .$\displaystyle r \:=\:167.0292252 \:\approx\:\boxed{167}$ km/hr

$\displaystyle \cos A \:=\:\frac{167^2 + 175^2 - 40^2}{2(167)(175)} \:=\:0.973721129$

Hence: .$\displaystyle A \:=\:13.16426633 \:\approx\:13^o$

Therefore, the heading is: .$\displaystyle \boxed{N\,5^o\,W}$