C4 Vectors

BA = a - b = (3, 2, -1) - (5, 1, 1) = (-2, 1, -2)
|BA| = sqrt(2^2 + 1^2 + 2^2) = sqrt9 = 3

BC = c-b = (7, -1, 0) - (5, 1, 1) = (2, -2, -1)
|BC| = sqrt(2^2 + 2^2 + 1^2) = sqrt9 = 3

(BA).(BC) = -2(2) -2(1) -1(-2) = -4

cosX = (BA).(BC)/(|BA||BC|) = -4/(3x3) = -4/9 = 4/9 (same either way around, since cos graph reflects in y axis)



cosX = 4/9
Area = 1/2 absinC
if you draw a right angled triangle using the cos value you have, then work out the other side... does that make sense?
the other side = sqrt ( 9^2 - 4^2 ) = sqrt65 = sqrt5sqrt13
so sinX = O/H = sqrt65 / 9

Area = 1/2 absinC = 1/2 (3)(3)sqrt65/9 = (sqrt65)/2


AC = c-a = (7, -1, 0) - (3, 2, -1) = (4, -3, 1)
CD = d-c = (7, 0, 3) - (7, -1, 0) = (0, 1, 3)

(AC).(CD) = 4(0) -3(1) + 1(3) = 0
but
(AC).(CD) = |AC||CD|cosX
--> cosX = 0
--> X = 90
--> vectors are perpendicular.


AD = d-a = (7, 0, 3) - (3, 2, -1) = (4, -2, 4)
|AD| = sqrt ( 4^2 + 2^2 + 4^2) = sqrt36 = 6

DB = b-d = (5, 1, 1) - (7, 0, 3) = (-2, 1, -2)
|DB| = sqrt(2^2 + 1^2 + 2^2) = sqrt9 = 3

--> AD: DB = 6:3 = 2:1


I hope that makes sense... if any of it doesn't, yell at me and I'll try again Vectors are just bags of fun all the way...

you are good at explaining (5yrs later)

Thank you

How do you know which terms have dy/dx at the end when they are differentiated implicitly?

The terms where you differentiate y have a dy/dx on the end:

$y=y^2 + x^2$ differentiated becomes $\dfrac{dy}{dx} = 2y \dfrac{dy}{dx} + 2x$

y=x^(x^2), find dy/dx in terms of x.