Vectors Question

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February 16th 2013, 12:32 PM
SeerAlpha

Vectors Question

Hi I'm stuck on the following questions:

a) Find an equation to describe the set of points equidistant from A(2, -1, 3) and B(1, 2, -3).
b) Find the coordinates of two points that are equidistant from A and B.

Please show the steps and thanks for your help! :)
February 16th 2013, 12:39 PM
emakarov

Re: Vectors Question

(1) Write the (square of) distance from (x, y, z) to A(2, -1, 3).
(2) Write the (square of) distance from (x, y, z) to B(1, 2, -3).
(3) Equate (1) and (2).
February 16th 2013, 12:53 PM
Plato

Re: Vectors Question

Quote:

Originally Posted by SeerAlpha

Hi I'm stuck on the following questions:

a) Find an equation to describe the set of points equidistant from A(2, -1, 3) and B(1, 2, -3).
b) Find the coordinates of two points that are equidistant from A and B.

Here is a second way. Write the equation of the plane that contains the midpoint of $\overline{AB}$ that has normal $\overrightarrow {AB}$ .
February 16th 2013, 01:09 PM
SeerAlpha

Re: Vectors Question

The book's answer is x-3y+6z = 0 where P(x,y,z) is the point but I didn't get that as an answer... so did you get the book's answer or a different one? I probably missed something :)
February 16th 2013, 01:18 PM
HallsofIvy

Re: Vectors Question

What did you get and how did you get it? What is the point halfway between A and B? What is the vector $\vec{AB}$ ?
February 16th 2013, 01:31 PM
SeerAlpha

Re: Vectors Question

I did PA = (2, -1, 3) - (x,y,z)
= (2-x, -1-y, 3-z)

PB = (1,2,-3) - (x,y,z)
= (1-x, 2-y, -3-z)

|PA| = |PB|

sqrt[(2-x)^2 + (-1-y)^2 + (3-z)^2] = sqrt[(1-x)^2 + (2-y)^2 + (-3-z)^2]
4 - 4x + x^2 + 1 + 2y + y^2 + 9 - 6z + z^2 = 1 - 2x + x^2 + 4 - 4y + y^2 + 9 +6z + z^2
-2x + 6y - 12z = 0

(sorry i'm not sure how to add the direction arrows)
February 16th 2013, 01:42 PM
Plato

Re: Vectors Question

Quote:

Originally Posted by SeerAlpha

I did PA = (2, -1, 3) - (x,y,z)
= (2-x, -1-y, 3-z)

PB = (1,2,-3) - (x,y,z)
= (1-x, 2-y, -3-z)

|PA| = |PB|

sqrt[(2-x)^2 + (-1-y)^2 + (3-z)^2] = sqrt[(1-x)^2 + (2-y)^2 + (-3-z)^2]
4 - 4x + x^2 + 1 + 2y + y^2 + 9 - 6z + z^2 = 1 - 2x + x^2 + 4 - 4y + y^2 + 9 +6z + z^2
-2x + 6y - 12z = 0

That is that correct answer.
$-2x+6y-12=0$ is $x-3y+6z=0$ .
February 16th 2013, 01:46 PM
SeerAlpha

Re: Vectors Question

... Doh! Should of compared the two answers more carefully. Hmm I'm going to try part b but can you guys give me a hint on how to start it?
February 16th 2013, 02:02 PM
SeerAlpha

1 Attachment(s)

Re: Vectors Question

Oh and if possible I have one last question that I really need help with... Please see the attached image for the question and diagram. Attachment 27066
February 16th 2013, 02:16 PM
SeerAlpha

Re: Vectors Question

Ok I got part b just need some guidance on the diagram question.
February 16th 2013, 02:22 PM
emakarov

Re: Vectors Question

It is better to start a new thread.

You have coordinates of all vectors. Use the fact that $\overrightarrow{OP} \cdot\overrightarrow {AE}=|\overrightarrow{OP} |\cdot|\overrightarrow{AE}|\cos\alpha$ where $\alpha$ is the required angle, $\overrightarrow{OP} \cdot\overrightarrow{AE}$ is the dot product and $|\overrightarrow{OP} |$ is the length of $\overrightarrow{OP}$ , i.e., $\sqrt{\overrightarrow{OP}\cdot\overrightarrow{OP}}$ .
February 16th 2013, 02:50 PM
SeerAlpha

Re: Vectors Question

I think I solved it. You've all been a great help! :)