# Vectors

• Feb 1st 2013, 06:43 AM
Tutu
Vectors
Hi I'm not sure if this is the right forum, but I apologize if it is not.

I am very new to vectors and am very confused by it, would really appreciate the help!

1.) Determine the nature of triangle ABC using distances: A(0,0,3), B(2,8,1) and C(-9,6,18)
For the lengths, I got AB to be sqrt(24) units
I got BC to be sqrt(414) units
and AC to be sqrt(373) units
So it is not an isoceles or equitorial triangle.

Then I did the dot product
I didn't get zero for all of them..
I had taken AB dot BC, BC dot AC, AC dot AB, wondering if it is right. If it is convenient, can someone explain why you dot like that to me, I do not understand it.

2.) A sphere has a centre C(-1, 2, 4) and diameter [AB] where A is (-2,1,3). Find the coordinates of B and the radius of the circle.

What does [AB] mean! And how do I even do this..

Thank you so much!
• Feb 1st 2013, 09:20 AM
Soroban
Re: Vectors
Hello, Tutu!

You made some errors in the first one.
I'll correct them.

Quote:

1) Determine the nature of triangle ABC with vertices: A(0,0,3), B(2,8,1) and C(-9,6,18)

For the lengths, I got AB to be sqrt(72) units
I got BC to be sqrt(414) units
and AC to be sqrt(342) units
So it is not an isoceles or equilateral triangle. . Right!

Given two vectors $\vec u$ and $\vec v,\,$ if $\vec u \cdot \vec v \,=\,0,\,\text{ then }\,\vec u \perp \vec v.$

We have: . $\begin{Bmatrix}\overrightarrow{AB} \:=\:\langle 2,8,\text{-}2\rangle \\ \overrightarrow{AC } \:=\:\langle \text{-}9,6,15\rangle}\end{Bmatrix}$

Then: . $\overrightarrow{AB}\cdot\overrightarrow{AC} \:=\:\langle 2,8,\text{-}2\rangle \cdot \langle \text{-}9,6,15\rangle \:=\:\text{-}18 + 48 - 30 \:=\:0$

Hence: . $AB \perp AC.$

Therefore, $\Delta ABC$ is a right triangle.

Quote:

2) A sphere has a centre C(-1, 2, 4) and diameter AB where A is (-2,1,3).
Find the coordinates of B and the radius of the sphere.

Did you make a sketch?
Code:

      A    r    C    r    B       * - - - - - * - - - - - *   (-2,1,3)    (-1,2,4)    (x,y,z)
Going from A to C, we move:
. . +1 in the x-direction, +1 in the y-direction, and +1 in the z-direction.

Going from C to B, we make the same moves:
. . $\begin{Bmatrix}x &=&\text{-}1+1 &=& 0 \\ y&=&2+1&=&3 \\ z&=&4+1&=&5\end{Bmatrix}$

Therefore: . $B(0,3,5)$

$r \:=\:AC\:=\:\sqrt{(\text{-}1-[\text{-}2])^2 + (2-1)^2 + (4-3)^2} \:=\:\sqrt{1+1+1}$

Therefore: . $r \:=\:\sqrt{3}$

• Feb 1st 2013, 08:20 PM
Tutu
Re: Vectors
Thank you so much!
May I ask, if ABC is to be a straight line, what am I supposed to get? I know if ABC is to be right angled, the AB will be perpndicular to AC, or BC will be perpendicular to CA but I got different values for AB, BC and AC so ABC is not isoceles or equilateral, and AC dot AB, BC dot CA did not give me 0.

The points are A(1,0,-3), B(2,2,0), C(4,6,6)

Thank you!
• Feb 2nd 2013, 08:37 AM
Soroban
Re: Vectors
Hello again, Tutu!

Quote:

If $ABC$ is a straight line, what am I supposed to get?

The points are: $A(1,0,\text{-}3),\:B(2,2,0),\:C(4,6,6)$

We certainly do not get a triangle.
(Well, maybe a very flat one.)

$\begin{Bmatrix}\overrightarrow{AB} &=& \langle1,2,3\rangle \\ \overrightarrow{BC} &=& \langle 2,4,6\rangle &=& 2\langle 1,2,3\rangle \\ \overrightarrow{AC} &=& \langle 3,6,9\rangle &=& 3\langle 1,2,3\rangle \end{Bmatrix}$

The vectors are multiple of the same angle; they are parallel.