# Vector calculus question

• March 13th 2013, 11:29 AM
FelixHelix
Vector calculus question
Hi - I'm totally stuck with this question: how to interpret it and tackle it. Any advice woiuld be greatly received!! We've not covered anything like this in classes...

Let

$A = \left( x_{A},y_{A},z_{A} \right)$
$B = \left( x_{B},y_{B},z_{B} \right)$

be two given distinct points in the Euclidean space. By finding the cartesian equation, descibe the surface representing the location of points M which are solutions of the equation

$\vec{AM}.\vec{MB} = 0$

Thanks, FH
• March 13th 2013, 11:50 AM
MINOANMAN
Re: Vector calculus question
It is easy
in the plane this equation represents a circle.
in the 3d it will be a sphere. Put M(x,y,z) find the vectors AM and MB and get their scalar product....the rest is easy.
• March 13th 2013, 12:07 PM
FelixHelix
Re: Vector calculus question
OK. So ive taken the scalar prodcut and get:
$\left( X_{M}.X_{A}-X_{B}.X_{A}-X_{M}.X_{M} \right)+\left( Y_{M}.Y_{A}-Y_{B}.Y_{A}-Y_{M}.Y_{M} \right)+\left( Z_{M}.Z_{A}-Z_{B}.Z_{A}-Z_{M}.Z_{M} \right)$

But what next?
• March 13th 2013, 01:03 PM
HallsofIvy
Re: Vector calculus question
No, that's incorrect. Taking the ordinates of point M to be $(x, y, z)$, the vectors AM and BM are $(x_a- x, y_a- y, z_a- z)$ and $(x- x_b, y- y_b, z- b_z)$ respectively. The dot product is $(x_a- x)(x- x_b)+ (y_a- y)(y- y_b)+ (z_a- z)(z- z_b)= -x_ax_b- (x_a+ x_b)x- x^2- y_ay_b- (y_a+ y_b)y- y^2- z_az_b- (z_a+ z_b)z- z^2= 0$.
• March 13th 2013, 02:09 PM
FelixHelix
Re: Vector calculus question
Quote:

Originally Posted by HallsofIvy
No, that's incorrect. Taking the ordinates of point M to be $(x, y, z)$, the vectors AM and BM are $(x_a- x, y_a- y, z_a- z)$ and $(x- x_b, y- y_b, z- b_z)$ respectively. The dot product is $(x_a- x)(x- x_b)+ (y_a- y)(y- y_b)+ (z_a- z)(z- z_b)= -x_ax_b- (x_a+ x_b)x- x^2- y_ay_b- (y_a+ y_b)y- y^2- z_az_b- (z_a+ z_b)z- z^2= 0$.

How do you get the sign to be negative for the terms $- (x_a+ x_b)x, - (y_a+ y_b)y & -(z_a+ z_b)z$ ?? Thats not how the above parenthesis expands?
• March 13th 2013, 03:55 PM
Soroban
Re: Vector calculus question
Hello, FelixHelix!

Quote:

Let: $\begin{Bmatrix}A &=& \left( x_a,y_a,z_a\right) \\ B &=& \left(x_b,y_b,z_b\right)\end{Bmatrix}$ be two given distinct points in Euclidean space.

By finding the cartesian equation, descibe the surface representing the location
of points $M$ which are solutions of the equation: $\overrightarrow{AM}\cdot\overrightarrow{BM}\,=\,0$

Let $M \,=\,(x,y,z)$

Then: . $\begin{Bmatrix}\overrightarrow{AM} &=& \langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle \\ \overrightarrow{BM} &=& \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle \end{Bmatrix}$

We have: . $\langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle\cdot \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle \;=\;0$

. . $(x-x_a)(x-x_b) + (y-y_a)(y-y_b) + (z-z_a)(z-z_b) \;=\;0$

. . $x^2 - (x_a\!+\!x_b)x + x_ax_b + y^2 - (y_a\!+\!y_b)y + y_ay_b + z^2 - (z_a\!+\!z_b)z + z_az_b \;=\;0$

. . $x^2 - (x_a\!+\!x_b)x + y^2 - (y_a\!+\!y_b)y + z^2 - (z_a\!+\!z_b)z \;=\;-x_ax_b - y_ay_b - z_az_b$

Complete the square:

. . $x^2 - (x_a+x_b)x + (\tfrac{x_a+x_b}{2})^2 + y^2 - (y_a+b_b)y + (\tfrac{y_a+y_b}{2})^2 + z^2 - (z_a+z_b)z + (\tfrac{z_a+z_b}{2})^2$

. . . . . . $\;=\;(\tfrac{x_a+x_b}{2})^2 + (\tfrac{y_a+y_b}{2})^2 + (\tfrac{z_a+z_b}{2})^2 - x_ax_b - y_ay_b - z_az_b$

. . $\left(x-\tfrac{x_a+x_b}{2}\right)^2 + \left(y - \tfrac{y_a+y_b}{2}\right)^2 + \left(z - \tfrac{z_a+z_b}{2}\right)^2 \;=\;\left(\tfrac{x_a-x_b}{2}\right)^2 + \left(\tfrac{y_a-y_b}{2}\right)^2 + \left(\tfrac{z_a-z_b}{2}\right)^2$

The locus of M is a sphere with center $C\left(\frac{x_a+x_b}{2},\,\frac{y_a+y_b}{2},\, \frac{z_a+z_b}{2}\right)$

. . and radius: $r \:=\:\sqrt{\left(\frac{x_a-x_b}{2}\right)^2 + \left(\frac{y_a-y_b}{2}\right)^2 + \left(\frac{z_a-z_b}{2}\right)^2}$

• March 13th 2013, 05:49 PM
FelixHelix
Re: Vector calculus question
Quote:

Originally Posted by Soroban
Hello, FelixHelix!Let $M \,=\,(x,y,z)$Then: . $\begin{Bmatrix}\overrightarrow{AM} &=& \langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle \\ \overrightarrow{BM} &=& \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle \end{Bmatrix}$We have: . $\langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle\cdot \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle \;=\;0$. . $(x-x_a)(x-x_b) + (y-y_a)(y-y_b) + (z-z_a)(z-z_b) \;=\;0$. . $x^2 - (x_a\!+\!x_b)x + x_ax_b + y^2 - (y_a\!+\!y_b)y + y_ay_b + z^2 - (z_a\!+\!z_b)z + z_az_b \;=\;0$. . $x^2 - (x_a\!+\!x_b)x + y^2 - (y_a\!+\!y_b)y + z^2 - (z_a\!+\!z_b)z \;=\;-x_ax_b - y_ay_b - z_az_b$Complete the square:. . $x^2 - (x_a+x_b)x + (\tfrac{x_a+x_b}{2})^2 + y^2 - (y_a+b_b)y + (\tfrac{y_a+y_b}{2})^2 + z^2 - (z_a+z_b)z + (\tfrac{z_a+z_b}{2})^2$. . . . . . $\;=\;(\tfrac{x_a+x_b}{2})^2 + (\tfrac{y_a+y_b}{2})^2 + (\tfrac{z_a+z_b}{2})^2 - x_ax_b - y_ay_b - z_az_b$. . $\left(x-\tfrac{x_a+x_b}{2}\right)^2 + \left(y - \tfrac{y_a+y_b}{2}\right)^2 + \left(z - \tfrac{z_a+z_b}{2}\right)^2 \;=\;\left(\tfrac{x_a-x_b}{2}\right)^2 + \left(\tfrac{y_a-y_b}{2}\right)^2 + \left(\tfrac{z_a-z_b}{2}\right)^2$The locus of M is a sphere with center $C\left(\frac{x_a+x_b}{2},\,\frac{y_a+y_b}{2},\, \frac{z_a+z_b}{2}\right)$. . and radius: $r \:=\:\sqrt{\left(\frac{x_a-x_b}{2}\right)^2 + \left(\frac{y_a-y_b}{2}\right)^2 + \left(\frac{z_a-z_b}{2}\right)^2}$

Then: . $\begin{Bmatrix}\overrightarrow{AM} &=& \langle x\!-\!x_a,\,y\!-\!y_a,\,z\!-\!z_a\rangle \\ \overrightarrow{BM} &=& \langle x\!-\!x_b,\,y\!-\!y_b,\,z\!-\!z_b\rangle \end{Bmatrix}$the vector isnt BM but MB. Will this matter?