Question

Vectors question

Answer 1

I LOVE VECTORS :)

Answer 2

there is the Q and the A. I don't understand how the answer is gotten to even after writing them all out in component form and so on. please help

Answer 3

that png is a statement, not a question

Answer 4

ahhh, first one is the question .... got it

Answer 5

\[r-a=<x-a1,y-a2,z-a3>\]\[r-b=<x-b1,y-b2,z-b3>\]------------------------------

dot:\[x^2-(a1+b1)x+a1b1\]\[y^2-(a2+b2)y+a2b2\]\[z^2-(a3+b3)z+a3b3\]equals zero

Answer 6

do some completing the squares

Answer 7

looking to get:\[(x-m)^2+(y-n)^2+(z-p)^2+(constants)=0\]

Answer 8

\[(x-\frac{a1+b1}{2})^2−(\frac{a1+b1}{2})^2+a1b1\]\[(y-\frac{a2+b2}{2})^2−(\frac{a2+b2}{2})^2+a2b2\]\[(z-\frac{a3+b3}{2})^2−(\frac{a3+b3}{2})^2+a3b3\]

Answer 9

from this you can pull out the center\[\frac{1}{2}(a1+b1),\frac{1}{2}(a2+b2),\frac{1}{2}(a3+b3)\]

Answer 10

and the radius is the sqrt of the opposite of the constants added up

Answer 11

\[−(\frac{a1+b1}{2})^2+a1b1−(\frac{a2+b2}{2})^2+a2b2−(\frac{a3+b3}{2})^2+a3b3\]
negate it to get
\[(\frac{a1+b1}{2})^2-a1b1+(\frac{a2+b2}{2})^2-a2b2+(\frac{a3+b3}{2})^2-a3b3\]
and simpify

\[\frac{1}{4}(a1+b1)^2-a1b1+\frac 14 (a2+b2)^2-a2b2+\frac 14{(a3+b3)^2}-a3b3\]
\[\frac{1}{4}((a1+b1)^2-4a1b1+(a2+b2)^2-4a2b2+(a3+b3)^2-4a3b3)\]
\[\frac{1}{4}((a1+b1)^2+(a2+b2)^2+(a3+b3)^2-4(a1b1+a2b2+a3b3))\]
if a dot b = 0, we are good to go

Answer 12

if you can think of a thrm or prove that\[(r-a)*(r-b)=0 \]is the same as\[a*b=0\]then the problem is resolved

Answer 13

There is an error in the algebra dealing with the constants, in fact it does just drop out as given in the answer eg

a1b1 - ((a1+b1)/2)^2 ->

[4a1b1 - (a1^2 +b1^2 +2a1b1) ]/4

2a1b1 - a1^2 -b1^ 2 all over 4

which is (a-b)^2 over 4 and the others similar equal to r^2

Answer 14

..... latex and albegraing just dont mix ;) thnx

Answer 15

You are either very patient or a super fast typer.....:-)