Posted by christina on Friday, September 13, 2013 at 11:05pm.
I have no idea how you were taught to calculate vectors.
For any two points P(a,b) and Q(c,d)
vector PQ = < c-a, d-b>
I got
vector AB = <2,5>
vector BC = <-7,-3) and
vector AC = <-5,2>
let angle CAB = Ø
then vector AC . vector AB = |AC| |AB| cosØ
<-5,2> . <2,5> = | <-5,2>| |<2,5>| cosØ
-10+10 = √29 √29 cosØ
cosØ = 0
Ø = 90°
well, that was obvious from looking at the two vectors, their dot product is zero, so they are perpendicular
for angle ABC
<-3,-7> . <-5,-2> = √58 √29 cos B
15+14 = √58√29cosB
cos B =.707106...
B = 45°
Well, how about that ?
Suppose we had taken the length of each vector
|AB = √29
|AC| = √29
|BC| = √58
so it is isosceles, and since
(√58) ^2 = (√29)^2 + (√29)^2
so it is also right-angled, as I showed in my first method.
I have no idea how you were taught to calculate vectors.
For any two points P(a,b) and Q(c,d)
vector PQ = < c-a, d-b>
I got
vector AB = <2,5>
vector BC = <-7,-3) and
vector AC = <-5,2>
let angle CAB = Ø
then vector AC . vector AB = |AC| |AB| cosØ
<-5,2> . <2,5> = | <-5,2>| |<2,5>| cosØ
-10+10 = √29 √29 cosØ
cosØ = 0
Ø = 90°
well, that was obvious from looking at the two vectors, their dot product is zero, so they are perpendicular
for angle ABC
<-3,-7> . <-5,-2> = √58 √29 cos B
15+14 = √58√29cosB
cos B =.707106...
B = 45°
Well, how about that ?
Suppose we had taken the length of each vector
|AB = √29
|AC| = √29
|BC| = √58
so it is isosceles, and since
(√58) ^2 = (√29)^2 + (√29)^2
so it is also right-angled, as I showed in my first method.
THANK YOU SO MUCH!