Thread: Triangle defined by vectors

There's a triangle ABC whose vertexes are connected to the point P and there are vectors PA->(8;24) and PB->(-4;19) and PC->(1;7).
How to find the vector AB->
The arrow "->" must be above the letters but I don't know how to write them so.

The vector AB is what you have to add to PA to get PB: that is, it's PB-PA.

There are essentially two types of vectors, position vectors and free vectors.
Position vectors are anchored at a point or origion, free vectors aren't.
The vectors in your problem tell you how to get to each verticee from the point P

The vector P->A is described as <8,24> or 8 units right & 24 units up from p
The vector P->B is described as <-4,19> or 4 units left & 19 units up from p

If P is the point (0,0) we can locate A by following the position vectors instructions on how to get there. It will have coordinates (8,24)

In the same way we can locate B by following the position vectors instructions on how to get there. It will have coordinates (-4, 19)

The vector A->B is the instructions <left/right, up/down> on how to get to B FROM A. If you have plotted the points correctly then you can count the squares remembering to include the signs (left is -ve in going across and down is -ve vertically)

Alternatively

A->B = (P->B) - (P->A)
A->B = <-4, 19> - <8, 24>
A->B = <-4-8, 19-24>
A->B = <-12, -5>

Check this agrees with your diagram. To get to B from A travel 12 left and 5 down.

Hope this helps.

Now I understand. This made it clear: "The vector P->A is described as <8,24> or 8 units right & 24 units up from p"
And there's no matter what's the point P, it can be (0;0) or (2;3) or ... etc. Thanks.