243.12.5.27 - Verify vector identiy

[karush]

$\textsf{Write a complete solution.}\\$
$\textit{ok Let A be the point (1,2,3) and O the origin (0, 0, 0). }\\$
$\textit{Consider the points P (x, y, z) such that AP · OP − OA · OP = 2 − |OA|2.}\\$
$\textit{Show that the set of all such points is a sphere, and find its center and radius.}\\$

ok, will be dealing with this in the morning
but would like some suggestions?

 

[MarkFL]

Re: 243.12.5.27

I would begin by expressing the given directed line segments in component form:

\(\displaystyle \vec{AP}=\left\langle x-1,y-2,z-3 \right\rangle\)

\(\displaystyle \vec{OP}=\left\langle x,y,z \right\rangle\)

\(\displaystyle \vec{OA}=\left\langle 1,2,3 \right\rangle\)

And then:

\(\displaystyle \left|\vec{OA}\right|=\sqrt{1^2+2^2+3^2}=\sqrt{14}\)

Now, consider the definition of the dot product:

\(\displaystyle \text{a}\cdot\text{b}=\sum_{k=1}^n\left(a_kb_k\right)\)

Can you proceed?

 

[MarkFL]

Re: 243.12.5.27

Are you certain the given equation isn't:

\(\displaystyle \vec{AP}\cdot\vec{OP}-\vec{OA}\cdot\vec{OP}=2-\left|\vec{OA}\right|^2\) ?

On a side note, can you generalize such that $O$ and $A$ are arbitrary distinct points in the space?

 

[karush]

Re: 243.12.5.27

View attachment 7333

$\textsf{the dot product $u\cdot v$ of vectors $u=\langle u_1, u_2, u_3 \rangle$}$
$\textsf{and $v=\langle u_1, u_2, u_3 \rangle$ is}$
$$u\cdot v=u_1v_1+u_2v_2+u_3v_3$$

 

[MarkFL]

Re: 243.12.5.27

I suspected that vector magnitude was to be squared...makes for a nicer problem. ;)

 

[karush]

Re: 243.12.5.27

$\displaystyle \vec{AP}\cdot\vec{OP}-\vec{OA}\cdot\vec{OP}=2-\left|\vec{OA}\right|^2$
so since
$\displaystyle \vec{AP}=\left\langle x-1,y-2,z-3 \right\rangle$
$\displaystyle \vec{OP}=\left\langle x,y,z \right\rangle$
$\displaystyle \vec{OA}=\left\langle 1,2,3 \right\rangle$
then
\begin{align}
&\, \, \, \, \, \, \left\langle x-1,y-2,z-3 \right\rangle \cdot \left\langle x,y,z \right\rangle
-\left\langle 1,2,3 \right\rangle \cdot \left\langle x,y,z \right\rangle\\
&=[(x-1)x+(y-1)y+(z-3)z]-[x+2y+3z]\\
&=x^2-x+y^2-y+z^2-3z-x-2y-3z\\
&1^2-1+2^2-2+3^2-3(3)-1-2(2)-3(3)=\color{red}{ -12}
\end{align}
and also
\begin{align}
2-\left|\vec{OA}\right|^2=2-14=\color{red}{ -12}
\end{align}
the standard equation of a sphere is
$(x - h)^2 + (y - j)^2 + (z - k)^2 = r^2$
$\textbf{so |OP|^2=14=radius}$
$\textit{not sure about the radius??}$

 

[MarkFL]

Re: 243.12.5.27

Let's look at where you should have:

\(\displaystyle x^2-2x+y^2-4y+z^2-6z=-12\)

Hint: Try completing the square on $x,\,y,\,z$. To better appreciate this problem, I do recommend exploring the more general problem I spoke of in post #3.

 

[karush]

Re: 243.12.5.27

Let's look at where you should have:

\(\displaystyle x^2-2x+y^2-4y+z^2-6z=-12\)

Hint: Try completing the square on $x,\,y,\,z$. To better appreciate this problem, I do recommend exploring the more general problem I spoke of in post #3.

\begin{align*}\displaystyle
&x^2-2x+y^2-4y+z^2-6z=-12\\
&x^2-2x+1+y^2-4y+4+z^2-6z×9\\
&=-12+1+4+9=2\\
&(x-1)^2+(y-2)^2+(z-3)^2\\
&=2\\
&=(\sqrt{2})^2
\end{align*}

radius=2

 

[MarkFL]

Re: 243.12.5.27

\begin{align*}\displaystyle
x^2-2x+y^2-4y+z^2-6z&=-12\\
x^2-2x+1+y^2-4y+4+z^2-6z×9&=-12+1+4+9=2\\
(x-1)^2+(y-2)^2+(z-3)^2&=2=(\sqrt{x})^2
\end{align*}

so is $\sqrt{2}$ the radius?

Yes, and I assume this is just a typo, but you want:

\(\displaystyle (x-1)^2+(y-2)^2+(z-3)^2=2=\left(\sqrt{2}\right)^2\)

Now, suppose we define:

\(\displaystyle \left(x_A,y_A,z_A\right)\) = point $A$

\(\displaystyle \left(x_O,y_O,z_O\right)\) = point $O$

And so we have:

\(\displaystyle \vec{AP}=\left\langle x-x_A,y-y_A,z-z_A \right\rangle\)

\(\displaystyle \vec{OP}=\left\langle x-x_O,y-y_O,z-z_O \right\rangle\)

\(\displaystyle \vec{OA}=\left\langle x_A-x_O,y_A-y_O,z_A-z_O \right\rangle\)

And then:

\(\displaystyle \left|\vec{OA}\right|=\sqrt{\left(x_A-x_O\right)^2+\left(y_A-y_O\right)^2+\left(z_A-z_O\right)^2}\)

Now suppose the given equation is:

\(\displaystyle \vec{AP}\cdot\vec{OP}-\vec{OA}\cdot\vec{OP}=k-\left|\vec{OA}\right|^2\) where $0<k$

We then find:

\(\displaystyle \left(\left(x-x_A\right)\left(x-x_O\right)+\left(y-y_A\right)\left(y-y_O\right)+\left(z-z_A\right)\left(z-z_O\right)\right)-\left(\left(x_A-x_O\right)\left(x-x_O\right)+\left(y_A-y_O\right)\left(y-y_O\right)+\left(z_A-z_O\right)\left(z-z_O\right)\right)=k-\left(\left(x_A-x_O\right)^2+\left(y_A-y_O\right)^2+\left(z_A-z_O\right)^2\right)\)

After some algebra, this becomes:

\(\displaystyle \left(x-\left(x_A+x_O\right)\right)^2+\left(y-\left(y_A+y_O\right)\right)^2+\left(z-\left(z_A+z_O\right)\right)^2=\left(\sqrt{k}\right)^2\)

And hence, the given equation describes a sphere with center \(\displaystyle \left(x_A+x_O,y_A+y_O,z_A+z_O\right)\) and radius $\sqrt{k}$. :)