help-36

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[void needle]

i would like to know the concept of linear equations am reallynew tp it

[polar wyvern]

what kind of equations?

[void needle]

a linear equation maybe explain a basic start up

[polar wyvern]

well graphically these equations represent straight lines

[summer sedge]

They are in the form $y = mx + b$

[soft zealot]

Aà

[ornate tree]

Linear = straight line

[polar wyvern]

like with basic y=x represent a line equally inclined in x and y axis

[void needle]

ohh

like representing on a graph

[summer sedge]

That's a use for them, yes

[polar wyvern]

yes

[void needle]

so i am gonna send u a question

then u help me out

[polar wyvern]

ok

[summer sedge]

!da2a

[final saddle]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[void needle]

yes

that one

[summer sedge]

Ok so

[late rose]

These aren't linear equations since they involve more than two variables

[void needle]

kk

so wat type are they

[summer sedge]

They're linear in a 3d space

I don't know what the term is

[void needle]

i think thye are linear algebric equations

[late rose]

Has your school taught you Gaussian elimination yet?

[void needle]

noo

[late rose]

okay you can still solve it without

[void needle]

maybe i an try

[late rose]

so you wanna use one equation, isolate a variable (e.g. x_1), and then substitute it into the other two equations

[void needle]

ok

[late rose]

Let me know when you're done

[void needle]

mind if i illustrate with another question

[late rose]

no problem

[sand kindle]

You can use matrix 3x3

find det

And get roots

By Cramer’s rule

Because matrix is square delta exists

Or just use Gauss’s method

As you wish

@void needle

[final saddle]

@void needle Has your question been resolved?

[void needle]

thats the one 0

have u seen it

[final saddle]

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[void needle]

.reaopen

have u seen it

[late rose]

can you .reopen it

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[calm oxide]

Hi, can someone explain this question to me?

[mild iris]

is the answer c ?

@calm oxide

[calm oxide]

I have no idea

theres no answer sheet

[mild iris]

its just a guess that there shouldnt be any restrictions upon what a and b could be so R

cause in the complex plane its i -i , and then real and negative real axis

no boundations to it

and it just says a set of complex numbers so yea

ntg crazy about it ig

[calm oxide]

sorry im a bitlost

[mild iris]

umn

[calm oxide]

oh wait

i get it

[mild iris]

complex numbers are defined by a+bi right

[mild iris]

u sure ?

i can try in a bit easier way if u want

[calm oxide]

nah i get it

[calm oxide]

yes

it had to be c

[mild iris]

yeaa

are you okayish at physics by any chance ?

if yes could you look at my doubt aswell ?

[calm oxide]

sure

.close

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[leaden roost]

I need help

[leaden roost]

with a question

pls

I need to learn how to graph f(x) = 2(x+5)^3 -4 Into a graph
the class im taking is advanced functions grade 12, the unit is polynomal functions

[vital crag]

first plot x^3

then shift and scale x^3 using the table

,tex .transformation rules

[leaden roost]

wait waitng

[soft zealot]

riemann

[leaden roost]

how to plot X^3

[vital crag]

,w plot y = x^3

[soft zealot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rie.mann

[leaden roost]

but the thing is

hmm

ok then what

[vital crag]

did you see this

[leaden roost]

oh ok

shift and scale

how to do that

[vital crag]

did you see this table

[leaden roost]

yes

but idk where to start

[vital crag]

try just plotting (x+5)^3

[vital crag]

find the correct transformation type

[final saddle]

@leaden roost Has your question been resolved?

[iron cliff]

do you still need help

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[late geyser]

I'm doing this problem of simplifying complex fractions I literally don't know how to do this I'm currently trying to multiply by ab

[whole halo]

what did you get

[late geyser]

a^3+b^3 / a^2-ab+b^2

oh whoops

wrong question thats my bad i meant to upload this one

[whole halo]

remember that multiplying by ab changes the value of the fraction

you need to do something else so that the fraction retains the same value

[late geyser]

I'm trying to get rid of the fraction so wouldnt i multiply by the LCD

[whole halo]

lets think about what youre doing here

if you want to simplify 0.5 / 0.2,

are you saying that simplifies to 5 / 0.2?

shouldnt they simplify to 5/2 instead?

[late geyser]

yes

[whole halo]

so what step do you do to go from 0.5/0.2 to 5/2?

[late geyser]

multiplied both by 10

[whole halo]

so how would you do this correctly?

[late geyser]

umm lemme think

make the numerator a(a^2)+b(b^2)

[late geyser]

was this correct in the question

[whole halo]

youre forgetting something again

lets try this slower

[whole halo]

what step did you do here?

[late geyser]

multiplied by ab

[whole halo]

so what step are you forgetting?

[late geyser]

multiplying by ab in the denominator

[whole halo]

there you go

,,\frac{\frac{a^2}b+\frac{b^2}a}{a^2-ab+b^2}=\frac{a^3+b^3}{ab(a^2-ab+b^2)}

[soft zealot]

mtt

[late geyser]

is it simplified?

[whole halo]

of course not

what next?

[late geyser]

distribute the ab

[whole halo]

no

the next step needs you to recognize something about a^3 + b^3

and also about a^2 - ab + b^2

[late geyser]

oh sum of 2 cubes

(a+b)(a^2-ab+b^2)/ab(a^2-ab+b^2)

[edgy mauve]

Now u can cancel out

The 2nd term

[late geyser]

a+b/ab

[whole halo]

now you can divide the fraction

to get a simplified version

[late geyser]

wdym how do you simplify a+b/ab

i thought thats it

[whole halo]

do you know how to divide by a variable?

[late geyser]

?

[whole halo]

yes or no?

you need to divide by ab

[late geyser]

no idk

[whole halo]

lets say you had (5 + 3) / (5 * 3)

can you divide this

[late geyser]

do 8/15

[whole halo]

now suppose you cant do 5 + 3

what would you do instead

[late geyser]

5+3/15

[whole halo]

then?

[late geyser]

1/3+1/5?

[whole halo]

there you go

(a + b)/(ab) = 1/b + 1/a

[late geyser]

so a/ab+b/ab?

[whole halo]

yea

[elfin marsh]

Damn what level of math is this

Is this hs 2 or 3

[late geyser]

this is for my pre calc homework

[elfin marsh]

Oh that’s why

I’m doing hs 2a and b

Rn

[late geyser]

technically college precalc but i take it at my high school

[elfin marsh]

And for this year imma do 2ab 3a Pre calc

[elfin marsh]

Damn I needa lock in

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[final saddle]

Please don't occupy multiple help channels.

[slate narwhal]

<@&268886789983436800>

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[tranquil pine]

hello

[signal adder]

Hi

[slate narwhal]

got a question?

[slate narwhal]

hi, checking again for questions

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@tranquil pine Has your question been resolved?

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[lethal totem]

!da2a

[final saddle]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[slate narwhal]

please just send your question! it's faster that way

let me know when you've sent it. i'll pin it for others

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[tranquil pine]

oh yes

[tranquil pine]

im tired of this problem man

can someone please help me understand it

@slate narwhal

[glossy zephyr]

I think i can help if you want

[tranquil pine]

please

[glossy zephyr]

ill try to walk you through the line of thought involved on solving this problem

So, basically, we know the weight force created by the light itself

We have to balance it by an opposite force created by the three cables

[tranquil pine]

yes 27

pounds

[tranquil pine]

well dont they want the force on each cable

[glossy zephyr]

Im more accostumed to metric system, so i wont be using much of the actual numbers but just the logic

[glossy zephyr]

Yeah, im just going over the steps on how to find it

[tranquil pine]

okay

well yeah the cables are pulling back

but i just dont understand after this point

[glossy zephyr]

Since you know that the three cables are evenly spaced, the forces that are not parallel to the weight are cancelled out, right?

Aka, it doesnt move side to side because of the cables

This means that you can kinda ignore the fact that this is xyz, and instead just consider a single xy force

[tranquil pine]

yes there are horizontal forces pushing along eachother

so they cancel

[tranquil pine]

but what does this have to do with the canceling of the horizontal forces

[glossy zephyr]

You can just consider one of the forces over the plane it forms to the "center point"

[tranquil pine]

i see

[glossy zephyr]

Then you can decompose that single force

Since you know that the horizontal doesnt really matter because between the three theyll cancel out

You can just evaluate for the vertical

and multiply it by 3 (because of three cables)

[tranquil pine]

wait wait

im getting lost

so they are asking for tention in terms of L

right

by tension they mean

magnitude correct?

[glossy zephyr]

yeah

[tranquil pine]

okay can you scale that from L

like find the components

or does it not work like that

[glossy zephyr]

yes, but:

Thing is, the angle formed between the cable(black) and the light(blue) depends on the length of the cable,

For that you can just basically do trigonometry and define it as a function

[tranquil pine]

i dont understand

[glossy zephyr]

Lemme draw two examples

You can see that the left cable is obviously longer, and it has a bigger angle than the one in the right***.

Based on the length of the cable, we can find the corresponding angle

which will we use to find the vertical component of the force

[tranquil pine]

so basically magnitude -> is found from veritcal and horizontal components -> vertical component depends on what theta is, -> so we need to find theta

[glossy zephyr]

Yeah

You basically solve it the opposite way as its usually done

Want me to help with how the function would look like?

[tranquil pine]

i see, and so we need an actual value for theta right

[tranquil pine]

which will follow this right

[tranquil pine]

thjis]

this*

[glossy zephyr]

yeah

Theta is a function of length

Vertical component needed is a function of theta

The actual tension is a function of the vertical component.

[tranquil pine]

i dont get this

the function of part

all i know so far is what we need to find

the missing peices

[glossy zephyr]

You know that the down force of the light is 27 pounds right?

[tranquil pine]

yes

[glossy zephyr]

We have 3 cables, so each cable has to be doing a tension force which Vertical Component has to be 27/3 to balance it out, right?

[tranquil pine]

why is it vertical

[glossy zephyr]

horizontal we ignore

[tranquil pine]

and not on the cable

like why is it not diagnla

[glossy zephyr]

When working with forces you have the advantage of being able to find the actual force using any of the components and the angle

You know how the horizontal component is = Force . cos \theta ?

Well, Force = Horizontal Component / Cos \theta

Same goes for the vertical

Its basically an intermediate step

[tranquil pine]

cos(theta) = x/force => force * cos(theta) = x

[glossy zephyr]

or force = x / cos(theta)

as long as you know theta (thats why we are interested in theta to start with), we can just translate between each other

[tranquil pine]

right

[glossy zephyr]

Want me to guide you through all the steps of the "whole function" they ask for? T(L)

[tranquil pine]

yes please but i have a question, force = x / cos(theta) is supposed to give us theta that we need to find the vertical component?

[glossy zephyr]

the opposite, knowing theta, we can find the force based on the vertical component, ill try to make sense of it later down the line

[tranquil pine]

okay so

im trying to peice it

[glossy zephyr]

We have to define a function that will allow us to find the angle based on L

[tranquil pine]

we need magnitude -> get vertical component -> get theta to find vertical component

[glossy zephyr]

the magnitude comes later, again, try to follow with me how the full function should be pieced together

[tranquil pine]

yes but im saying we need to work on finding theta right?

to go to magnitude at the end

as a result

theta -> gives us vertical component -> gives us magnitude

[glossy zephyr]

i hope you remember about trig, but you can derive the function from SOHCAHTOA

[tranquil pine]

yes

i know the trig

[glossy zephyr]

you have any clue which trig function and how the formula would end up looking like?

[tranquil pine]

we need theta

so we can do

vertical = sin right

[glossy zephyr]

Ill spoil it, we need to relate the two sides we "know" to the angle, since we know the adjacent, and our variable occupies the hypotenuse, we use

cos(theta) = 18/L

again, L is just like an x

[tranquil pine]

why cos

why this operation specifically as a whole

why

[glossy zephyr]

to find the angle

[tranquil pine]

okay why not sin

or tan

[glossy zephyr]

tan relates the opposite and adjecent

sin relates the opposite and hypotenuse

We only know the adjacent and hypotenuse

[tranquil pine]

hypotnuse is L tho

its as unknown as h

[glossy zephyr]

-> We are defining a function on L

[tranquil pine]

but i thought we were focused on theta for now?

[glossy zephyr]

Theta is the result of said function, you know inverse trig?

if you want it simplified, you can do
cos(theta) = 18/L ->
theta = arccos(18/L)

[tranquil pine]

yes

but why is L relevant

[tranquil pine]

yes but why is it relevant to find theta in itself

[glossy zephyr]

To be able to find the real magnitude of the tension and not just its vertical component

[tranquil pine]

okay you lost me

hee

here

[glossy zephyr]

We know the vertical has to be 27/3,
since multiplied by 3 we have 27, which is what we need to balance the -27 from the weight

[tranquil pine]

why is it veritcal like that

specifically

[glossy zephyr]

we got 27 pounds excerting force downwards, so the three cables have to be excerting 27 pounds force upwards

that means that the sum of their forces is equal to 27 in the vertical axis, and since they are equally distributed, each one has 27/3

[tranquil pine]

i understand but arent the cables pulling diagnally

[glossy zephyr]

remember we ignore the horizontal

cause they cancel each other out

[tranquil pine]

yes but

the cables are pulling in a telted line

[glossy zephyr]

I just stated that their vertical component is 27/3

From there i have to find what is the actual magnitude by using the angle

[tranquil pine]

can you please go over this part again

[glossy zephyr]

Lets go over how force balance works

[tranquil pine]

okay

[glossy zephyr]

For an object to be static the sum of all the forces has to be 0, right?

[tranquil pine]

yes

correct

[glossy zephyr]

In this situation, we have a total of 4 forces

The weight, which is pointing straight down

And three tensions, which we know a few things of:

They are equally distributed, they have the same magnitude and they are diagonal, all pointing to the axis in which the light is fixed

[tranquil pine]

yes

[glossy zephyr]

We can infere a few thing already by know,

Since all 3 tensions are equally distributed and have the same magnitude, their horizontal forces have to cancel out.

in simpler words, the lamp doesnt move around

[tranquil pine]

correct

i agree on this

but then when it comes to the whole 9 and cos thing i get lost

[glossy zephyr]

Then, if we can ignore the horizontal, the only part which we care about the force is the vertical

[tranquil pine]

yes

[glossy zephyr]

going about the specifics:

If we know the lamp weighs 27 pounds, the sum of the three verticals has to be also 27

if it wasnt, it would fall or rise

[tranquil pine]

wait bro, wouldnt we ignore the horizontal if we were looking at all forces combined

but if we were to look at only 1

it isnt canceling with anything right

[glossy zephyr]

We are cancelling because when re-applying to the three it wouldnt really matter if there was or not.

[tranquil pine]

wdym by reapplying to the 3

[glossy zephyr]

imagine we do find the horizontal on a single cable

When we apply that horizontal to all three (Which we already stated that they are equally distributed)
They will all add to 0

So we just simply skip them

[tranquil pine]

i see now

yeah even when we are looking at just one

in reality its being canceled with as were looking at it

right

its not like were cutting it from the whole part

and looking at it

by cutting i mean physically

[glossy zephyr]

Going back, since we know their magnitude is equal, and they are equally distributed, we know they have a vertical of 27/3 (9) each, since 9 * 3 = 27 which is what we need as opposing upwards force

[tranquil pine]

yes..

its the vertical component because they are pulling up?

okay

so i agree so far

theres a 3 pound force on each cable

vertically

and we need magnitude at the end

so we need vertical component

which is 3

wait 9 sorry

9

[glossy zephyr]

So basically this right?

[tranquil pine]

right

[glossy zephyr]

I didnt drew the horizontal cause, again, we dont care

[tranquil pine]

yes

[glossy zephyr]

Well, analogous to how we can find the vertical component of a force using the angle

We can find the magnitude of the force using the angle and the vertical

in equation:

Fy = F . sinθ
->
F = Fy / sinθ

Like so, our next step is to find θ

[tranquil pine]

whats F_y

[glossy zephyr]

vertical component of the force / tension

mostly cause we usually use the y axis as the vertical

[tranquil pine]

didnt we find that to be 9 btw

[glossy zephyr]

Yes, but we want to find F

Using Fy

and θ

[tranquil pine]

wait so h =9?

but h was the vertical component

[glossy zephyr]

Thats only in forces, not in the true measurement scale

[tranquil pine]

wait what

[glossy zephyr]

You could have a cable thats 200 inches long applying a tension of 10 pounds

a 10 pound force doesnt mean that the cable is 10 inches long

[tranquil pine]

oh

[glossy zephyr]

ill recap:
We know
F = Fy / sinθ

We know Fy, we have to find θ

once we do its basically solved

[tranquil pine]

im kind of stuck on the 9 vs h

[glossy zephyr]

what do you think is h?

[tranquil pine]

i get 9 is the force and h is the distance but i dont understand the fuller picture

like when we think of components we think of forces right

9 is a force going up

so whats the issue

[glossy zephyr]

9 is the vertical component of the force*
h we dont really care for since its just a height

which doesnt serve any purpose for the problem

[tranquil pine]

interesting

so we found our vertical component basically?

[glossy zephyr]

Fy = 9 = vertical component

Thats basically a given by the problem

[tranquil pine]

but now its weird cause we have a vertical component but then we have an 18 inch radius

which isnt a force component

so how are we gonna work with a side that is a component and one that iosnt

[glossy zephyr]

Youre mixing the physical space with the force diagram

The only thing they share is the angle

[tranquil pine]

wait so let me recap, we want magnitude, -> we need vertical component for that, but we just found it

so isnt that <0, 9> (even though i know this is wrong)

earlier theta would determine the vertical component so we needed to find theta,,,, but now we found the vertical component

[glossy zephyr]

yes, the thing is, we cant find the real magnitude without the angle

[tranquil pine]

but dont we have the vertical component 9 and the horizontal one 0?

[glossy zephyr]

the fact that the sum of horizontals is 0 doesnt mean that themselves are 0

[tranquil pine]

i thought they cancel eachother in the background

[glossy zephyr]

I think this might help

I didnt drew the third one cause it would be confusing

The grey line in the middle would be the third cable, but the idea is the same

[tranquil pine]

yes so we have vertical component

and horizontal

[glossy zephyr]

No, we dont have the horizontal

[tranquil pine]

but its being canceled out with

the rest

[glossy zephyr]

the horizontal force could be 1 gazillion but since they cancel themselves, their sum is equal to 0

So we dont have any info beyond that their sum is 0

We do have on their vertical components, thats why we need the angle

[tranquil pine]

ohh okay but i thougth it was 9

so why does angle matter now

i thought we initially needed to the angle to find the veritcal component

[glossy zephyr]

Since we wanna know F
and know Fy = 9
We just need θ now.

[tranquil pine]

wait what do we need theta for then

[glossy zephyr]

F = Fy / sinθ

Theta is the angle

[tranquil pine]

what do we need angle for

[glossy zephyr]

You remember how to find the vertical and horizontal components of a force?

[tranquil pine]

not the reaction 😂

[tranquil pine]

but we have it though

usually youd use trig

but we have 9 already

[glossy zephyr]

Do you remember how to do it tho?

[tranquil pine]

well yeah

for vertical its y = sin(theta) = h/L

[glossy zephyr]

Not really

You forgot the fact you need to multiply by the force itself to get out the component

[tranquil pine]

uh

h = L * sin(theta)

@glossy zephyr

[glossy zephyr]

Thats the height, which, again, we dont really care for

We are working with forces here

[tranquil pine]

L = h /sin(theta)?

@glossy zephyr

[tranquil pine]

okay yeah

but why are we doing this

i thought we initially had the plan to : find magnitude -> from vertical component -> from theta

what im confused on:

we said theta determines the vertical component, but then we have it as 9, but also we have h

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

<@&286206848099549185>

[slate narwhal]

.close (OP has moved to a new channel)

[final saddle]

Channel closed

Closed by @slate narwhal

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[late gazelle]

could someone give a definition of composition of matrixes, in some cases its acts as addition and in others as multiplication and im confused on where which is used

[severe canyon]

Can you send an example of both cases you mentioned?

Because for the addition I have some doubt

[final saddle]

@late gazelle Has your question been resolved?

[final saddle]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[scenic pendant]

$2^k \sum^{\infty}_{n=\text{max}(0,k)} 4^{-n}$

[soft zealot]

alee

[scenic pendant]

it is of the form

$\sum_{n=0}^{\infty} q^n$

[twin pivot]

Hi

[scenic pendant]

hi

[soft zealot]

alee

[scenic pendant]

<@&268886789983436800>

troll

[barren hound]

?

[scenic pendant]

hes trolling

[ornate tree]

Please claim your own channel

[twin pivot]

Sorry

[barren hound]

oh you're the 2+2 guy. stop trolling.

[scenic pendant]

<@&286206848099549185>

[inner atlas]

Take a look at how you can possibly rewrite the equation to get a n power by itself

[final saddle]

@scenic pendant Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

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[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[next thorn]

If the addition of two unit vectors (lets say a and b) is a unit vector, so then 2a-3b is a vector and its magnitude is:

[next thorn]

I dont know how to approach this.

I tried to draw it geometrically but- was either too hard or irrelevant.

[craggy plank]

Oh, new question

[next thorn]

there are two more

[ornate tree]

Could you translate that?

[next thorn]

my pleasure

you mean type it in arabic

[tired walrus]

so you kinda don't need to geometrize it

[next thorn]

اذا كن مجموع متجهي وحده هو متجه وحده فائن الفرق بين ضعف الاول و ثلاثه امثال الثاني هو متجه معياره...

[tired walrus]

but you need either the angle between a and b, or the value of a•b

[ornate tree]

اذا كن مجموع متجهي وحده هو متجه وحده فائن الفرق بين ضعف الاول و ثلاثه امثال الثاني هو متجه معياره...
If the sum of two vectors is a vector, then the difference between twice the first and three times the second is a vector whose magnitude is...

[ornate tree]

no no, I mean the options

[next thorn]

7
sqrt(7)
19
sqrt(19)

[ornate tree]

Alright, you know the drill

what have you tried in advance?

just post everything regardless useful or not

:))

[tired walrus]

are you familiar w this

[next thorn]

drawing them out and then idk staring at it... but... since they're equal and a + b = c (which forms an equilateral triangle)

[next thorn]

shouldnt the angle be 120

[next thorn]

yes

very.

[ornate tree]

Look, the sum of two unit vectors are is also a unit vector, meaning that the angle between the two vectors should be greater than 90 degree, do you agree?

[tired walrus]

good, then find a•b from this

[next thorn]

ill do that in a second.

[next thorn]

who said cosine has to be negative

[ornate tree]

before moving on, would you like me to explain Ann's method or mine first?

Mine is more like geometric approach

and visualization

[next thorn]

ik where this came from, and from this, a ' b = -1/2 so cos(x)= -1/2 so the angle is indeed 120! but where does this take me anyway?

[tired walrus]

a•b = -1/2 good

now write out |2a-3b|^2 in the same fashion

[next thorn]

oh.

ok i see where this is going

[next thorn]

just to make sure, magn of 2a squared is 4 * magn of a squared

[tired walrus]

|2a|^2 = 4 |a|^2 = 4, yes

[next thorn]

|2a-3b|^2 = 14 + 12cos(x)

i... hope?

idk the calculations were messy.

[tired walrus]

4+9=13 not 14

[next thorn]

oh.

right

[tired walrus]

and now put -1/2 for cos(x)

[next thorn]

7

sqrt(7) it is

[tired walrus]

yes

[next thorn]

okay yes i see

ty

@ornate tree

sorry for the ping

[ornate tree]

what's up

[next thorn]

can you explain your geometric appraoch now

[ornate tree]

Certainly

[next thorn]

lovely

[ornate tree]

So let's say $\vec a + \vec b = \vec c$

[soft zealot]

This is sad 😢

[next thorn]

wheres vector b

[ornate tree]

ah mb

fumbling

[next thorn]

no problem 🤩

[ornate tree]

The statement that a, b, and c are unit vectors still holds

is it okay so far?

[next thorn]

yes

a+b=c

yes.

[ornate tree]

Can you draw me $2\vec a - 3\vec b$?

[soft zealot]

This is sad 😢

[next thorn]

oh this is what i tried but i fumbled the drawig too hard

okay i can try again

[ornate tree]

Feel free to post your work when your stuck

[next thorn]

l

it took a while 😔

[next thorn]

also here i said that 2a-3b is like 2a+ (-3b)

[ornate tree]

lemme check

[ornate tree]

yes, it is correct

[next thorn]

🤩

now what.

[ornate tree]

Now can you find me the magnitude of the green vector, which is the sum of 2a and -3b?

Hint: Cosine law

@next thorn

[next thorn]

mm

[ornate tree]

need more guidance?

[next thorn]

give me a bit

[ornate tree]

Alright

what's the angle between a and b again?

[next thorn]

120?

[ornate tree]

yes

how about 2a and b?

[next thorn]

ok hm question

[next thorn]

also 120

[ornate tree]

Decent

how about a and -b?

[next thorn]

i assume 60?

[ornate tree]

yep

[next thorn]

so that angle is 60

[ornate tree]

this one? no

[next thorn]

yes

yes

okay so

[ornate tree]

a and b >> 120
2a and b >> 120
2a and -b >> 60
2a and -3b >> 60

[ornate tree]

and be careful because they don't begin at the same spot

[next thorn]

cosine rule thats that cos(120) = ( 4|a|^2 + 9|b|^2 - g^2)/(2|a| |b|)

and

that is

cos(120) = (4+9-g^2)/2

forgot a square

and g is the magn of

green

[ornate tree]

$\cos(120 degree) = ( 4|a|^2 + 9|b|^2 - g^2)/(2|a| |b|)$

[next thorn]

yes

14

wait what

no

4+9=13?

(13-g^2)/2 = -1/2

whats r

[soft zealot]

This is sad 😢

[next thorn]

13- g^2=-1

[ornate tree]

nvm, I'll just set it as degree.

[next thorn]

-14

14

7

sqrt(7)

yes perfect

wait

what

-1 - 13 = -14

i need my cuaclatoro

what went wrong

[next thorn]

-1/2 = (13-g^2)/2

multiply both sides by two

13-g^2 = -1

how is it sqrt(14)

[ornate tree]

lemme check

[next thorn]

btw if you scroll up and down while looking at this the lines will start moving 🤩

.

.

.

.

.

.

sky are you okay 😭

[ornate tree]

yeah, I'm checking

oh bruh

I found the problem

[next thorn]

m?

[next thorn]

not /

[ornate tree]

Those guy should be 2a and -3b

[next thorn]

or wiat no

nvm

or wait what

ok yeah yeah

yes

i see now

-1/2 = 13-g^2/-12

sqrt(7)

yes amazing

[ornate tree]

Anything else?

[next thorn]

ty maam! there are two more questions

do i send them here or

[ornate tree]

Yeah, go ahead

[next thorn]

open another channel

[ornate tree]

I'll pin them as well

[ornate tree]

nah, I have pin permission

[next thorn]

can you ban people

[ornate tree]

no 😭

[craggy plank]

Skill issue

[next thorn]

not me sending you the question in arabic and expecting you to answer 🤩

if |a|=2 , |b|=3 , |c|=12, and they're all perpendicular on eachother like- how the x y and z planes are aligned

[next thorn]

whats |a + b + c|

[ornate tree]

if |a|=2 , |b|=3 , |c|=12, and they're all perpendicular on eachother like- how the x y and z planes are aligned
whats |a + b + c|?

[ornate tree]

Alright, let the fun begin

What have you tried?

[next thorn]

let me try what i tried again

i wish life was as simple as (|a +b + c| = |a| + |b| + |c|

[ornate tree]

LOL, same

[next thorn]

hold on, i still wanna keep trying

[ornate tree]

Yes, take your time

no hurries

Don't forget to ping me when you're done

[next thorn]

@ornate tree hints 😭 ?

[ornate tree]

yeah sure, I have two approaches. Algebric and Geometric

pick one first

[next thorn]

algebric

[ornate tree]

Sure

Let's assume
a = (2, 0, 0)
b = (0, 3, 0)
c = (0, 0, 12)

[next thorn]

WHGATHS

[ornate tree]

They're vectors btw

[next thorn]

yes.

[ornate tree]

Need more hints?

[next thorn]

no

[ornate tree]

they've given the magnitude, so I add them to it

[next thorn]

I really hope the answer is not just

magn of (2 , 3 , 1 2)

bc id cry

first of all

i have a question

[next thorn]

what justifies the zeroes here? i mean it can be anything

as long as the magn is equivalent

[ornate tree]

Of course, but it satisfies the question's requirement.

[next thorn]

.

[ornate tree]

You just need to prove this case is vaild, the rest will be vaild as well

[ornate tree]

yeah you probably will LOL

[next thorn]

interesting

yeah

okay

i thought of this approach

but then i said

it doesnt have to be exactly like the x y and z planes

like it doesnt have to have 0s

[next thorn]

but this is like

saying that

a b c can be an infinite amount of vectors

so this applies to ALL vectors

of magn 2 3 and 12

and are perpendicular on eachother

[ornate tree]

it doesn't certainly. But your answer is only and it must fulfill every possible scenario.

Regardless of what the actual vectors are, the patterm looks the same, right?

[next thorn]

what do you meaaaaaaan every possible scenario

what scenarios

[next thorn]

no!

[ornate tree]

how come?

[next thorn]

(72 , 72 , 0)

can be

the one with magn 12

cant it?

ok nv

i fucked smth up

[ornate tree]

no no, I'm referring to the patten
Like the shape they form

[next thorn]

yes

they all look like x y z

no matter their mags

[ornate tree]

exactly, they look like a corner of a cube

[next thorn]

alright

i get your point

but

[next thorn]

did you just make an xyz plane out of ear pickers

lovely.

[thin cloud]

-# peak

[edgy mauve]

This looks really cool for some reason

Satisfactory

[edgy mauve]

In these case r u allowed to use algebra
As in (a+b+c)² formula

[next thorn]

i thought

of that

toooooooooooooo

but like

how?

[edgy mauve]

Sooo does it work

[next thorn]

whats the formula gonna be like

theres 3 letters

are we gonna assume that a+b=k

[thin cloud]

(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

[edgy mauve]

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

[thin cloud]

that works

[edgy mauve]

There is a formula for tha5

[next thorn]

i didnt learn that. at all.

[edgy mauve]

If u want to derive then yes

[edgy mauve]

Ik it is derivable from whatever u learnt

[next thorn]

so we can keep on adding letters forever

and that formula will still work

[edgy mauve]

It wont be forever

[thin cloud]

no sadly

[edgy mauve]

It will

[next thorn]

forever as in like

idk twenty letters

[craggy plank]

[next thorn]

thats forever for me

[edgy mauve]

Sure haha

[craggy plank]

I’ll let them cook for now. Give me some time to come up with a method to explain that.

[next thorn]

Im just

not convinced

on

why

can we use x-y-z

for any

3 perpendicular

vectors

[craggy plank]

Yeah, and I’m responsible for that. I should find some examples to show my statement is correct.

[next thorn]

I'd appreciate that a lot.

[craggy plank]

I’ll go have some dinner first. Will DM you if I find one.

[thin cloud]

we can't it just looks like what x-y-z axis is

[next thorn]

so?

[next thorn]

take your time.

theres a fourth question

last one

and that calls for a new channel ig?

[edgy mauve]

Wasnt 20 terms rlly

[next thorn]

close enough.

ok yes

[edgy mauve]

It's okay to share here too ig

[next thorn]

i see how that is

[thin cloud]

do you know how to add 2 vector?

[next thorn]

i dont think i would ever compute that mid exam

[next thorn]

supposedly.

yes

[edgy mauve]

Well it's a handy formula at times

[thin cloud]

could you draw 2 perpendicular vector and its resultant vector

[next thorn]

I hate english so mych

why are letters so hard to draw

[thin cloud]

yes, okay now let call vec(a+b) vector c

[edgy mauve]

Uk cross product of two vectors ,could represent the other one

[next thorn]

ignore my c

[next thorn]

wdym?

cross product is the magn of one times the projection of the other,

if thats what you mean

wait a minte

wait

no

no ono nono

i didnt learn cross product yet

[edgy mauve]

U r mixing it with dot product ig

[next thorn]

idk what that is

[edgy mauve]

Vector product?

[next thorn]

yes, im supposed to learn it today

[next thorn]

yes

[thin cloud]

oh wait

[edgy mauve]

Oh then , it should be ignored for now

[thin cloud]

let call that vector d, not to confused with the vector c given

[next thorn]

okay..

[thin cloud]

so sum of vector a and b is vector d

a+b+c= d + c

[next thorn]

yes

[thin cloud]

now the red line and green line is also perpendicular

do you understand that part?

vector d and vector c are perpendicular

[next thorn]

yes

they are

am i supposed to be catching smth rn?

[thin cloud]

a+b+c become c+d

[next thorn]

mhm

[thin cloud]

do you know how to find resultant vector of c and d?

[next thorn]

you mean to draw it? yes its gonna go outward from their start points

[thin cloud]

yup, what about its magnitude

sorry my lap is on low batttery

[next thorn]

sqrt( mag(d)^2 + mag(c)^2 )

[thin cloud]

yes

what about mag(d)?

mag(c) is given

[next thorn]

mag(d) can come from magn a and magn b

using that square formula

[thin cloud]

yeah

that's it

[next thorn]

oh i see

thats so- long

because

in my exams

each question is supposed to take max 3 mins

[thin cloud]

it's $\sqrt{a^2+b^2+c^2}$

[soft zealot]

Alexis_Fx

[next thorn]

are those the mags

[thin cloud]

not that long if you know the approach

[thin cloud]

yes

[next thorn]

ok let me

try to

recomprehend

[thin cloud]

if vec a,b,c are mutually perpendicular

[next thorn]

wha you said

[next thorn]

oh

yes

mhm

okay let just try to fully grasp what you did

[thin cloud]

okay so the idea is , you've learnt how to find mag of sum of 2 vector

So if there're 3 or more

like a+b+c+...

we just find resultant vector of a and b

then find resultant of that vector and vec c

and repeat

if you have the coordinate, use coord

If not, do it geometrically like i did

[ornate tree]

exactly

[thin cloud]

You're k, right?

[ornate tree]

make another guess 🙂
In the hlounge

I'll read them

[thin cloud]

take too much brain cell to make a guess, let just keep it secret

[next thorn]

the resultant of a , b , c, its start point lets called (x , y ,z) and its end point is (x_2 , y_2 , z_2) and yes, indeed, x_2 - x = 2
y_2 - y = 3
z_2 - z = 12

so yeah

i get it now

the start point can be whatever, its just that when its reduced from the end point of the resultant vector, it will always give us the mag of x-y-z coordinates

[thin cloud]

that's not the way I did,do you understand why x_2-x=2

[next thorn]

yes

but you

set the foundation

or made me think about it

starting from your approach

and yes it makes a lot of sense

so ty

to everyone, really.

[thin cloud]

okay sure, if they're not mutually perpendicular you can't you it

[next thorn]

yes

[thin cloud]

okay great

[next thorn]

I can also make sense of that

no one told me linear algebra would be this bitchy

i HATE it.

anyways, last question

[ornate tree]

decent

wait, did Alexis explain why my approach works?

[next thorn]

alexis explained the resultant of a b c

[thin cloud]

no... he seems to be firgured it out himself

[next thorn]

ye

[ornate tree]

oh, I'll take that as yes :))

[ornate tree]

Even better

[next thorn]

lemme translate rq

[ornate tree]

Alright, I'll pin the translation

[next thorn]

a b c is a triangle with d, halving BC.
if |ab + ac| = |ab - ac|, whats the measure of the angle B A C

first, let me try to do this myself again

[ornate tree]

Pinned, ping me or Alexis when you're done

oh damn....I accidentally leaked my identity

[thin cloud]

-# the background remind me of James Doaks

[final saddle]

@next thorn Has your question been resolved?

[next thorn]

@craggy plank @thin cloud is it 90

i approached this algebrically

squaring both sides

and using the square formula

2 a b cos(theta) = - 2 a b cos(theta)

cos(theta) = -cos(theta)

thats... ninety.

[thin cloud]

It's 90

[next thorn]

yaaay

i finally did smth by myself

[thin cloud]

xD, there're a better way to do it

[next thorn]

oh pelase elaborate

[thin cloud]

give me a min

[next thorn]

my brother in islam (probably), do you not know that witch stuff is haram?

[thin cloud]

yk how we use parallelogram yo find the resultant vector

ABDC here's a parallelogram

[next thorn]

thats an ugly parallelogram no offense

[next thorn]

yes

[thin cloud]

AB+AC=AD and AB-AC=BC

from the given AD = BC

[next thorn]

yes

[next thorn]

yes

[thin cloud]

and a parallelogram with the diagonal segment are equal is...

A rectangle

so BAC=90 degree

[next thorn]

oh so it wasnt an ugly parallelogram

[next thorn]

yes

that makes so much sense

I do not like this field of math. too many possible ways to approach and yet i usually get none.

[thin cloud]

it's called geometry and that's why people hate geo

[next thorn]

how much time do you need to get good at visulaizing these stuff

[thin cloud]

a lots 3b1b videos

[next thorn]

oh yes i need to watch that

[ornate tree]

what's that?

[next thorn]

3blue1brown

youtuber

[ornate tree]

I see

[next thorn]

ok im done

now im going to learn cross product

i see a negative positive sign this is not gong to be okay

[thin cloud]

also I learnt drawing in perspective and a bit 3d

[thin cloud]

gl

[next thorn]

tym alexis and sky

I wish i could pay you back

but im broke.

[ornate tree]

I barely participate in the last hour lol

[next thorn]

do you guys perhaps want a idk transistion metals tutor

[next thorn]

what about the last 5 💀

[ornate tree]

yeah....anything else? Sorry I was at another channel so I didn't respond

[next thorn]

nope! ty for everything maam

have an amazing day

[ornate tree]

you too

[next thorn]

.close

[final saddle]

Channel closed

Closed by @next thorn

Use .reopen if this was a mistake.

[ornate tree]

[plush mulch]

It's just arithmetic

[next thorn]

all i see is arabic i cant read = devil stuff

i also dont see how that's arithmetic

[plush mulch]

Just because you can't read that means it's devil? Come on dude

[latent dragon]

Dude, seriously?

[plush mulch]

Man it's just arithmetic

I don't know how "counting" would end up being devil stuff, lmao

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[novel jackal]

Im supposed to simplify the left side to the form on the right with the addition of y < x