Given a math example of: "A person walks 3km east, then 1.5km south west. What is the resultant displacement?"
The study guide provides these workings;
A displacement of 3km east has x -component 3, y -component 0.
A displacement of 1.5km south west has: $$x -component -1.5 cos 45 = > -1.06$$ $$y -component -1.5 cos 45 = -1.06$$
The resultant has; $$x -component 3 - 1.06 = 1.94$$ $$y -component 0 -1.06 = -1.06$$
Why did the 3km displacement not get calculated similar to the 1.5?
example: $$3cos(45) + (-1.5)cos(45) = 2.12 + (-1.06) = 1.06$$
$1.06$ instead of $1.94$. Does this have something to do with the y component of the 3km displacement being a $0$?
EDIT:
I tried Arthur's suggestion with a following example but it doesn't work?
A plane with airspeed 200km/hour heads due north. A 100 km/hour north east wind (coming from the north east) begins to blow.
What is the resultant velocity of the plane relative to the ground?
We use symbols:
vPA = velocity of plane relative to air
vAG = velocity of air relative to ground
vPG = velocity of plane relative to ground = PA AG
$$vPA : x - component: 0$$ $$y -component: 200$$ $$AG v : x - component - 100 cos 45 = -70.7$$ $$y - component - 100 cos 45 = -70.7$$
It's definitely something to do with the $0$ component right? I just don't know the rule or reasoning behind it to use it on everything else.
It did get calculated the same way: $$ x: \quad 3\cdot \cos0=3\\ y: \quad 3\cdot\sin 0=0 $$ It is, however, almost trivial to see immediately from the drawing (and the text) that this is the result; going straight eastward will only ever change the $x$-component, and the change in the $x$-component must be equal to the distance walked. So they skipped the calculations in the study guide.