Question

PLEASE HELP

Consider the three points P1 : (1; 1; 0), P2 : (6; 2; 1), and P3 : (4; 3; 2).
(a) [1 marks] Find the displacement vector from P1 to P2.
(b) [1 marks] Find the displacement vector from P1 to P3.
(c) [4 marks] Find the angle between the two vectors found in parts
(a) and (b).

Answer 1

1. P1 - P2
2. P1 - P3
\[3. \cos \theta = \frac{ ||a||\,\,||b|| }{ a\cdot b }\text{ where }a\cdot b \text{ is the inner product}\]

Answer 2

so with p1-p2 would it be lthe answer be(1-6, 1-2, 0-1)?

Answer 3

P1-P2 = (-5, -1, -1)

Answer 4

wait but isnt rule that you go p2-pi because you go final minus initial?

Answer 5

"displacement vector from P1 to P2"

Answer 6

yeah so p2 is the final point and p1 is the initial point. P1 to P2

Answer 7

yeah it would be... to go from 1 to 6 you have to add 5, etc...

Answer 8

oh sorry I had typed the question in wrong. it was -6 in P2 and -2 in P3. My bad

Answer 9

it okay... i flipped the formula for cos...\[\cos \theta =\frac{ a\cdot b }{ ||a||\,\,||b|| }\]

Answer 10

the formula you had originally wsa right because a.b = |a||||b|| cosθ

Answer 11

yes but solve for cos

Answer 12

man i am out of it today

Answer 13

no... the last is correct not the first

Answer 14

Stupid mistake thanks for your help

Answer 15

me too!

Answer 16

I get it

Answer 17

sleep

Answer 18

ha will do

Answer 19

for me!

Answer 20

woops well sleep for the both of us then

Answer 21

lol