Looks to me like
AB + BC + CA = 0
Similarly for the other one. Getting back to A, the sum is just 0 again.
The vectors form a closed polygon, so the net result is zero.
These are vectors as well:
(a) AB + __ = CA = 0
(b) AB +BC + CD + DA = __
the first was a mistake i wanted these ones. Complete these and please if you can, explain.
4 answers
If you want to use vector notation, then if the vector to A is a, and to B is b, etc.
Then
AB = b-a
BC = c-b
CA = a-c
AB+BC+CA = b-a+c-b+a-c = 0
AB is b-a because if there is a vector u such that a+u=b, then u=b-a.
Then
AB = b-a
BC = c-b
CA = a-c
AB+BC+CA = b-a+c-b+a-c = 0
AB is b-a because if there is a vector u such that a+u=b, then u=b-a.
Thanks this will help.
I only knew of:
AB + BC = AC(triangle rule)
and
AB + BC + CD = AC + CD = AD(parallelgram rule)
I only knew of:
AB + BC = AC(triangle rule)
and
AB + BC + CD = AC + CD = AD(parallelgram rule)
Suppose 4ABC has vertices A(−6, −2), B(−8, 0) and C(−4, 0):
Find AB, BC and CA. Show that their sum is a zero vector.
Find AB, BC and CA. Show that their sum is a zero vector.