How do you draw a line that is perpendicular to 3 axes?

how do you draw a line that is perpendicular to 3 axes?

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en.wikipedia.org/wiki/Seven-dimensional_cross_product
youtube.com/watch?v=0t4aKJuKP0Q
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Physics majors everyone!

Not possible

wtf i hate math now

Easy, just draw it into the 4th dimension

but the axis for that dimension IS that line. where do you put it???

en.wikipedia.org/wiki/Seven-dimensional_cross_product

>how do you draw a 3 dimensional line

WTF you smoking OP?

nigga a line in 3 dimensions is just a linear combination of the 3 axes. if it's perpendicular to all 3, it can't be a linear combination, it has to be linearly independent. what are YOU smoking???

Make a 4x4 matrix. The top row has the 4 axis unit vectors, and the next three have the components of the 3 vectors. Take the determinant.

I'm assuming this extends to arbitrary nxn but I'm not going to try to prove it at 3am.

Nevermind, the proof is really easy. It works for n-1 vectors in n-space.

You don't.
Not difficult to work out the properties of higher-dimensional spaces -- just add another coordinate -- but they can't be drawn.
Any number of YouTube videos where they've animated projections of 4D objects into 3D space. Only other possibilities are leaving out one of the "regular" axes or using a property (like color) to indicate a 4th coordinate.

define perpendicular

cross product

Perpendicular to the others.

animations can be drawn.

Easy, draw a line. It is perpendicular to all dimensions except for the one that it is in. As long as we can accept that there are at least 4 dimensions, any line that you draw will always be perpendicular to 3.

by not drawing the other axes

How do you draw a line that is perpendicular to 2 axes? A piece of paper only has 2 axes

not possible to physically make something such that this is true, but it could be represented on paper or in a 3d model, albeit hard to read

Easy! Draw any old line, then draw little parallelograms at the intersect between the line and the three axes, label them all as 'a' and define a as "equal to 90 degrees or pi/4 radians"

Since we can use perspective to make a plane look like a 3 dimensional space, could it be possible to use a hologram to simulate a 4 dimensional space ?

no, cross product is a binary operator or vectors.
Im looking for a rule of assignment
[math]\varphi:V\times V \rightarrow \{0,1\}[/math]

>Since we can use perspective to make a plane look like a 3 dimensional space, could it be possible to use a hologram to simulate a 4 dimensional space ?
You would still just use a plane to make it look like a 4 dimensional space. A hologram wouldn't add anything because our eyes can only see one side of it at a time anyway.
youtube.com/watch?v=0t4aKJuKP0Q
>inb4 "b-but it doesn't count!"
>Everything is an approximation, nothing we perceive is anything more than a questionable bag of hacky workarounds, 4D perception doesn't need to be anywhere close to perfect to "count," it just needs to be based in the display of objects as they would appear were they following 4D physics, and that's something computers can handle pretty easily, you just add an extra number to the arrays of attributes defining your shapes

dot product equals 0
a*b = a_x*b_x + a_y*b_y + a_z*b_z = |a|*|b|*cos(theta)

>you just add an extra number to the arrays of attributes defining your shapes

this is how brainlets think

inner product is more general and zero for orthogonal vectors is true for complex vector spaces too

did it

You're welcome

like this

your puny brain was only build to comprehend 3 dimensions.

Anime style

On a different piece of paper. Make sure to note "4th Dimension" at the top.

I don't see a 90-degree angle in there, son...

what the fuck are you talking about, faggot

because you're a brainlet