Thread: Need help with vectos

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..

Hey Tutu.

(Hint: Try using Pythagoras' Theorem).

But how, I do not know the coordinates of the other end of B?

You know the height of B and you know the hypotenuse.

Use the fact that a^2 + b^2 + c^2 = d^2 where you know d = SQRT(14) and a = -1, b = 2. (I'm using a 3D version of the Pythagoras' Theorem).

Sorry, how did you get a^2 + b^2 + c^2 = d^2 ? And how did you get those values..?
Sorry

This is the form of the distance formula in three dimensions (as opposed to the one in two).

The hypotenuse is given in the question and the rest is found from the values in your diagram.

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!

You are right in that a = -1 and b = -1 and you need to find a value of c where d = SQRT(14).

I got c to be sqrt(12)?

That looks right.

But from there, how do I find the two coordinates on the y-axis?
Sorry that I still cannot get this!

Find the y-point +- SQRT(12) of the B's y-point (which is -1 according to the diagram).

So it's -1+sqrt(12) and -1-sqrt(12)?

Yes, that is the idea.

I found the answer sheet, it says the answer is (0,2,0) and (0.-4,0)!