Hi please help me with this! How do I do part b!

I thought to find the midpoint of B and the negative y-axis, and then equate the y-coordinate to sqrt(14) but that did not seem to be the way..

Results 1 to 15 of 18

- February 1st 2013, 08:41 PM #1

- Joined
- Nov 2012
- From
- Singapore
- Posts
- 180

- February 1st 2013, 09:05 PM #2

- Joined
- Sep 2012
- From
- Australia
- Posts
- 3,619
- Thanks
- 592

- February 1st 2013, 09:21 PM #3

- Joined
- Nov 2012
- From
- Singapore
- Posts
- 180

- February 1st 2013, 09:27 PM #4

- Joined
- Sep 2012
- From
- Australia
- Posts
- 3,619
- Thanks
- 592

- February 1st 2013, 10:17 PM #5

- Joined
- Nov 2012
- From
- Singapore
- Posts
- 180

- February 1st 2013, 10:21 PM #6

- Joined
- Sep 2012
- From
- Australia
- Posts
- 3,619
- Thanks
- 592

- February 1st 2013, 10:27 PM #7

- Joined
- Nov 2012
- From
- Singapore
- Posts
- 180

## Re: Need help with vectos

I just went to search online,

they said a is the left right distance, b is the front back distance and c is the up down distance.

I understand why a^2 + b^2 + c^2 = d^2 but then shouldn't a be -1, b be -1, and c be 2?

How did you know sqrt 14 is the distance? The more I look at the diagram, it looks like the midpoint!

- February 1st 2013, 10:33 PM #8

- Joined
- Sep 2012
- From
- Australia
- Posts
- 3,619
- Thanks
- 592

- February 1st 2013, 10:41 PM #9

- Joined
- Nov 2012
- From
- Singapore
- Posts
- 180

- February 1st 2013, 10:43 PM #10

- Joined
- Sep 2012
- From
- Australia
- Posts
- 3,619
- Thanks
- 592

- February 1st 2013, 10:48 PM #11

- Joined
- Nov 2012
- From
- Singapore
- Posts
- 180

- February 1st 2013, 10:50 PM #12

- Joined
- Sep 2012
- From
- Australia
- Posts
- 3,619
- Thanks
- 592

- February 1st 2013, 11:17 PM #13

- Joined
- Nov 2012
- From
- Singapore
- Posts
- 180

- February 1st 2013, 11:20 PM #14

- Joined
- Sep 2012
- From
- Australia
- Posts
- 3,619
- Thanks
- 592

- February 1st 2013, 11:27 PM #15

- Joined
- Nov 2012
- From
- Singapore
- Posts
- 180