These two problems are based on Vectors, dot product and distance for sphere.

scolon94

New member
Problem 1: Let S1 be a sphere centered at(0, 1, -3) with radius 1 and let S2 be a sphere centered at (3, 5, -9) with radius 2. Find the distance between the two spheres.

problem 2: Given three non-zero vectors v1, v2, v3 we say that they are mutually orthogonal when v1 dot v2= 0, v1 dot v3=0 , v2 dot v3=0. Sho wthat if v1+v2+v3 are mutually orthogonal then for any vector v we can write v= av1+ bv2+ cv3 for some real number a,b,c. (hint: Use the dot product on the equation v= av1+ bv2+cv3)

MarkFL

Staff member
Can you show us what you have tried or what your thoughts are on how to begin so our helpers know where you are stuck and how best to help?

By the way, I edited your thread title to remove the plea for help, since the implication that help is being sought after is made by the act of posting.

soroban

Well-known member

Hello, scolon94!

Did you make a sketch?

(1) Let S1 is a sphere with center (0, 1, -3) and radius 1.
S2 is a sphere with center (3, 5, -9) and radius 2.
Find the distance between the two spheres.
Code:
                                * * *
*           *
*             *               *
*       *        *                 *
*         *
1        *    2              *
*     *-----*-----*---------*         *
S1       d  *         S2        *
*         *
*       *        *                 *
*             *               *
*           *
* * *
The distance between the two centers is:

. . $$\overline{S_1S_2} \:=\:\sqrt{3-0)^2 + (5-1)^2 + (-9-[\text{-}3])^2}$$

. . . . $$=\;\sqrt{3^2+4^2 + 6^2} \;=\;\sqrt{61}$$

Therefore, the distance $$d$$ is . . .

scolon94

New member
Can you show us what you have tried or what your thoughts are on how to begin so our helpers know where you are stuck and how best to help?

By the way, I edited your thread title to remove the plea for help, since the implication that help is being sought after is made by the act of posting.
For the first problem I did distance formula. the sqrt ((x-xo) + (y-yo) + (z-zo)) as a result i got sqrt(16). I'm confused what to do after.

Problem#2 I really don't understand what to do whatsoever. since abc are numbers I can't apply dot product. I thought of dividing by v on both sides but it doesn't lead me anywhere.

Thank you

scolon94

New member

Hello, scolon94!

Did you make a sketch?

Code:
                                * * *
*           *
*             *               *
*       *        *                 *
*         *
1        *    2              *
*     *-----*-----*---------*         *
S1       d  *         S2        *
*         *
*       *        *                 *
*             *               *
*           *
* * *
The distance between the two centers is:

. . $$\overline{S_1S_2} \:=\:\sqrt{3-0)^2 + (5-1)^2 + (-9-[\text{-}3])^2}$$

. . . . $$=\;\sqrt{3^2+4^2 + 6^2} \;=\;\sqrt{61}$$

Therefore, the distance $$d$$ is . . .
This is exactly what I did, I did get sqrt(61). But my professor e-mailed me saying " This is the start of the problem. You need to find the distance between the spheres, not between the radii."

MarkFL

Staff member

This is exactly what I did, I did get sqrt(61). But my professor e-mailed me saying " This is the start of the problem. You need to find the distance between the spheres, not between the radii."
What must you subtract from the distance between the centers of the spheres to find the (minimal) distance between their surfaces ?

scolon94

New member

What must you subtract from the distance between the centers of the spheres to find the (minimal) distance between their surfaces ?
1 and 2. I get it now. thank you !

how about the second problem ?

MarkFL

Staff member
We are given the vector equation:

$$\displaystyle \textbf{v}=a\textbf{v}_1+b\textbf{v}_2+c\textbf{v}_3$$

Now dot each side with itself, and use the criterion for orthogonality in the expansion. What do you obtain?

scolon94

New member
We are given the vector equation:

$$\displaystyle \textbf{v}=a\textbf{v}_1+b\textbf{v}_2+c\textbf{v}_3$$

Now dot each side with itself, and use the criterion for orthogonality in the expansion. What do you obtain?
I'm not sure if this is the correct way.

va1*vb1 + va2*vb2 + va3*vb3 = av1*av1 + bv2*bv2 +cv3*cv3

Or can I put on the right side the norm of V ^2 because of one of the properties ?

MarkFL

Staff member
What I did was:

$$\displaystyle \textbf{v}\cdot\textbf{v}= \left(a\textbf{v}_1+b\textbf{v}_2+c\textbf{v}_3 \right) \cdot\left(a\textbf{v}_1+b\textbf{v}_2+c\textbf{v}_3 \right)$$

And after distributing on the right and making use of the mutual orthogonality of the 3 vectors there, I obtained:

$$\displaystyle |\textbf{v}|^2=a^2|\textbf{v}_1|^2+b^2|\textbf{v}_2|^2+c^2|\textbf{v}_3|^2$$

Next, I would suggest writing each magnitude in component form.

scolon94

New member
What I did was:

$$\displaystyle \textbf{v}\cdot\textbf{v}= \left(a\textbf{v}_1+b\textbf{v}_2+c\textbf{v}_3 \right) \cdot\left(a\textbf{v}_1+b\textbf{v}_2+c\textbf{v}_3 \right)$$

And after distributing on the right and making use of the mutual orthogonality of the 3 vectors there, I obtained:

$$\displaystyle |\textbf{v}|^2=a^2|\textbf{v}_1|^2+b^2|\textbf{v}_2|^2+c^2|\textbf{v}_3|^2$$

Next, I would suggest writing each magnitude in component form.
I get sqrt(v1^2+ v2^2+v3^2) = a^2 sqrt(V1^2+V2^2+V3^2) +b^2 sqrt(V1^2+V2^2+V3^2) + c^2 sqrt(V1^2+V2^2+V3^2)

MarkFL

Staff member
I get sqrt(v1^2+ v2^2+v3^2) = a^2 sqrt(V1^2+V2^2+V3^2) +b^2 sqrt(V1^2+V2^2+V3^2) + c^2 sqrt(V1^2+V2^2+V3^2)
The magnitudes are all squared, so you will have no square roots.

For the 3 mutually orthogonal vectors, let's define:

$$\displaystyle \textbf{v}_i=\left\langle x_i,y_i,z_i \right\rangle$$

And so:

$$\displaystyle |\textbf{v}_i|^2=x_i^2+y_i^2+z_i^2$$

Also from our originally given equation, we may state:

$$\displaystyle \textbf{v}=\left\langle ax_1+bx_2+cx_3,ay_1+by_2+cy_3,az_1+bz_2+cz_3 \right\rangle$$

Make sure you understand how I obtained the above component form using scalar distribution and vector addition before proceeding.

And so:

$$\displaystyle |\textbf{v}|^2=\left(ax_1+bx_2+cx_3 \right)^2+\left(ay_1+by_2+cy_3 \right)^2+\left(az_1+bz_2+cz_3 \right)^2$$

Now, after you use this, the equation will have many more terms on the left than the right, but subtract away what's common to both sides, and then you should be able to use the component form of the dot product, along with the mutual orthogonality to get $0=0$ as your final result, which will then prove what you needed to prove.