Math Help

[Canadianbacon17 0]

Does anyone know a website where you can draw a line or circle and the equation is given for you?

[tokes 0]

Does anyone know a website where you can draw a line or circle and the equation is given for you?

For a circle:

(x−a)²+(y−b)²=r²

center = (a,B)

radius = r

For a straight line:

y = ax+b, a = slope of line, b = vertical translation

You could input the straight line into a TI83+, etc, if you have one, but that's overly complicated. If you need to find the line tangent to a circle just take the derivative, equate it at whatever point you're finding the tangent at, use that slope to plug it into your y = ax + b, then sub in the values at that point to find out your constant b

On a side note, anyone writing AP tests next week? BC calculus, here I come :(

[seanmccann 3]

This is the closest I can think of http://gcalc.net/

[Eazy_b97 1]

Hoping someone can help me with a little Linear Algebra

"Given the two points P(-4, -4, -5) and Q(3, -1, 3) find the point R that is 1/6 of the way from Q to P."

The answer is

(11/6, -9/6, 10/6)

...but how?

[ktang 34]

Hoping someone can help me with a little Linear Algebra

"Given the two points P(-4, -4, -5) and Q(3, -1, 3) find the point R that is 1/6 of the way from Q to P."

The answer is

(11/6, -9/6, 10/6)

...but how?

The way I would do it: do each coordinate separately, since the x-y-z coordinate axes are orthogonal to each other.

For x: Qx is at 3, Px is at -4, difference is -7. 1/6 of that is -7/6. Add that to 3 (Qx), and you get 11/6.

For y: Qy is at -1, Py is at -4, difference is -3. 1/6 of this is -3/6. Add that to -1 (Qy), and you get -9/6.

For z: Qz is at 3, Pz is at -5, difference is -8. 1/6 of this is -8/6. Add that to 3 (Qz), and you get 10/6.

[Jason Harris 31]

Man, it's kind of depressing what you forget over the course of your life.... :(

[Eazy_b97 1]

You can help me with this one then :)

The goal of this question is to write a given vector v = (-7, 7, -3) as a sum v = v1 + v2, where v1 is parallel to another given vector w, and v2 is orthogonal to w = (-3, 0, -5).

Answer:

v1 = (-3.176, 0.0, -5.294), and

v2 = (-3.824, 7.0, 2.294).

...but why?

[Fire0nIce228 1]

You people are some of the smartest people i could ever know..damn i suck at math and you guys are cranking this stuff out over the internet!

[ktang 34]

You can help me with this one then :)

The goal of this question is to write a given vector v = (-7, 7, -3) as a sum v = v1 + v2, where v1 is parallel to another given vector w, and v2 is orthogonal to w = (-3, 0, -5).

Answer:

v1 = (-3.176, 0.0, -5.294), and

v2 = (-3.824, 7.0, 2.294).

...but why?

So, if I understand the question correctly, another way to state it is that

v1 is parallel to w

and

v2 is orthogonol to w.

---------------

If v2 is orthogonol to w, then the dot product of v2 and w should be 0. (if I remember the terminology correctly, and if the terminology has not changed in the eons since I did this in school)

(v2x * wx ) + (v2y * wy) + (v2z * wz) = 0

Substituting the known values for w,

v2x * (-3) + v2y * (0) + v2z * (-5) = 0

v2x * (-3) + v2z * (-5) = 0

v2x * (-3) = v2z * 5

v2x = v2z * 5 / (-3) <--- Save this for later use (i)

---------------

If v1 is parallel to w, then v1's components are all in proportion to w's.

So,

v1x / wx = v1y / wy = v1z / wz

Since wy = 0, we can assume that

v1y = 0 <--- Save this for later use (ii)

That leaves us with:

v1x / wx = v1z / wz

Substituting,

v1x / (-3) = v1z / (-5)

v1x = v1z * 3 / 5 <--- Save this for later use (iii)

---------------

v1 + v2 = (-7, 7, -3) gives us 3 sets of equations (for each component)

v1x + v2x = -7

v1y + v2y = 7

v1z + v2z = -3

From (ii) above, we know that v1y = 0, so v2y = 7.

Substituting from (i) and (iii) above for v1x and v1z, we can get these 2 equations so solve:

(v1z * 3 / 5) + (v2z * 5 / (-3)) = -7

v1z + v2z = -3

You can solve this as a 2x2 matrix (or whatever the newfangled term is)

3/5 -5/3 | -7

1 1 | -3

to get

v1z = -90/17 (approx -5.294)

v2z = 39/17 (approx 2.294)

Then, using (i) and (iii) again,

v1x = 3/5 * v1z = 3/5 * (-90/17) = -270/85 (approx -3.176)

v2x = -5/3 * v2z = -5/3 * (39/17) = -65/17 (approx -3.824)

[ktang 34]

Man, it's kind of depressing what you forget over the course of your life.... :(

I'd better remember this for a little while longer, because I will need to help my son with his homework...

[WAMPS19 0]

People shouldn't be doing math like this until they get into college. I didn't do Vector Algebra until my Junior Year in College. If you are an Engineer, wait until Differential Equations, that is the fun stuff!

[Sven 1]

You can help me with this one then :)

The goal of this question is to write a given vector v = (-7, 7, -3) as a sum v = v1 + v2, where v1 is parallel to another given vector w, and v2 is orthogonal to w = (-3, 0, -5).

Answer:

v1 = (-3.176, 0.0, -5.294), and

v2 = (-3.824, 7.0, 2.294).

...but why?

So, if I understand the question correctly, another way to state it is that

v1 is parallel to w

and

v2 is orthogonol to w.

---------------

If v2 is orthogonol to w, then the dot product of v2 and w should be 0. (if I remember the terminology correctly, and if the terminology has not changed in the eons since I did this in school)

(v2x * wx ) + (v2y * wy) + (v2z * wz) = 0

Substituting the known values for w,

v2x * (-3) + v2y * (0) + v2z * (-5) = 0

v2x * (-3) + v2z * (-5) = 0

v2x * (-3) = v2z * 5

v2x = v2z * 5 / (-3) <--- Save this for later use (i)

---------------

If v1 is parallel to w, then v1's components are all in proportion to w's.

So,

v1x / wx = v1y / wy = v1z / wz

Since wy = 0, we can assume that

v1y = 0 <--- Save this for later use (ii)

That leaves us with:

v1x / wx = v1z / wz

Substituting,

v1x / (-3) = v1z / (-5)

v1x = v1z * 3 / 5 <--- Save this for later use (iii)

---------------

v1 + v2 = (-7, 7, -3) gives us 3 sets of equations (for each component)

v1x + v2x = -7

v1y + v2y = 7

v1z + v2z = -3

From (ii) above, we know that v1y = 0, so v2y = 7.

Substituting from (i) and (iii) above for v1x and v1z, we can get these 2 equations so solve:

(v1z * 3 / 5) + (v2z * 5 / (-3)) = -7

v1z + v2z = -3

You can solve this as a 2x2 matrix (or whatever the newfangled term is)

3/5 -5/3 | -7

1 1 | -3

to get

v1z = -90/17 (approx -5.294)

v2z = 39/17 (approx 2.294)

Then, using (i) and (iii) again,

v1x = 3/5 * v1z = 3/5 * (-90/17) = -270/85 (approx -3.176)

v2x = -5/3 * v2z = -5/3 * (39/17) = -65/17 (approx -3.824)

You work for the NASA, right?

Looks like freakin' rocket-sience.

[ktang 34]

You can help me with this one then :)

The goal of this question is to write a given vector v = (-7, 7, -3) as a sum v = v1 + v2, where v1 is parallel to another given vector w, and v2 is orthogonal to w = (-3, 0, -5).

Answer:

v1 = (-3.176, 0.0, -5.294), and

v2 = (-3.824, 7.0, 2.294).

...but why?

So, if I understand the question correctly, another way to state it is that

v1 is parallel to w

and

v2 is orthogonol to w.

---------------

If v2 is orthogonol to w, then the dot product of v2 and w should be 0. (if I remember the terminology correctly, and if the terminology has not changed in the eons since I did this in school)

(v2x * wx ) + (v2y * wy) + (v2z * wz) = 0

Substituting the known values for w,

v2x * (-3) + v2y * (0) + v2z * (-5) = 0

v2x * (-3) + v2z * (-5) = 0

v2x * (-3) = v2z * 5

v2x = v2z * 5 / (-3) <--- Save this for later use (i)

---------------

If v1 is parallel to w, then v1's components are all in proportion to w's.

So,

v1x / wx = v1y / wy = v1z / wz

Since wy = 0, we can assume that

v1y = 0 <--- Save this for later use (ii)

That leaves us with:

v1x / wx = v1z / wz

Substituting,

v1x / (-3) = v1z / (-5)

v1x = v1z * 3 / 5 <--- Save this for later use (iii)

---------------

v1 + v2 = (-7, 7, -3) gives us 3 sets of equations (for each component)

v1x + v2x = -7

v1y + v2y = 7

v1z + v2z = -3

From (ii) above, we know that v1y = 0, so v2y = 7.

Substituting from (i) and (iii) above for v1x and v1z, we can get these 2 equations so solve:

(v1z * 3 / 5) + (v2z * 5 / (-3)) = -7

v1z + v2z = -3

You can solve this as a 2x2 matrix (or whatever the newfangled term is)

3/5 -5/3 | -7

1 1 | -3

to get

v1z = -90/17 (approx -5.294)

v2z = 39/17 (approx 2.294)

Then, using (i) and (iii) again,

v1x = 3/5 * v1z = 3/5 * (-90/17) = -270/85 (approx -3.176)

v2x = -5/3 * v2z = -5/3 * (39/17) = -65/17 (approx -3.824)

You work for the NASA, right?

Looks like freakin' rocket-sience.

No, I work for Raytheon. Raytheon. Unlike NASA, our rockets only have to go up and go "boom".

Linear Algebra is an important basis for higher-order math. My solutions look more complicated than they really are because I have to type them in.

[Fredrik 0]

Man, it's kind of depressing what you forget over the course of your life.... :(

:D

I second that!

[jjtt99 0]

Who would have thought this type discussion would ever exist at a hockey forum!

There was another crazy conversation about odor neutralizers a while back. It got really scientific...very impressive.

You never know what you're going to find at Mod!

By the way, what colour laces will help me skate backward faster ;-)

[Chadd 916]

You work for the NASA, right?

Looks like freakin' rocket-sience.

No, I work for Raytheon. Raytheon. Unlike NASA, our rockets only have to go up and go "boom".

Linear Algebra is an important basis for higher-order math. My solutions look more complicated than they really are because I have to type them in.

That's cold, nice to know we have a real-life rocket surgeon here on MSH. ;)

[WAMPS19 0]

I don't know if I would be posting that I work for Raytheon if I were him. Well known company with a big customer base.

[Eazy_b97 1]

Alright Guys, I am working on some more practice questions, hopefully I can get some more help on these.

#1) Given 3 Independant Vectors:

X1 X2 X3

260 -73 7

-247 66 7

44 -8 1

-52 18 0

0 2 0

From R (to the 5th), find a fourth Vector (X4) such that set (X1-X3) remains Linearly Independant

I will likely post some more later on.

[Shagel 0]

Man, it's kind of depressing what you forget over the course of your life....  :(

:D

I second that!

i also second that except im in grd 10 lol