Calculus 3 Tangent Vectors

Please help Veeky Forums I hate asking for help with this but I've been getting frustrated trying to figure out what this is asking for over 10 hours. I usually am a few sections ahead of the class, but I wasn't for yesterday's lecture. Can some kind user please point out what is being asked here and what strategy to solve this is?

What is this i, j, k bullshit? Why not just call it v=(7,1,a)?

It's the historical notation from Gibb's which was taken from Hamilton's quaternions which was taken from a+bi forms of complex numbers

But they're not quaternions though and it's really cumbersome notation.

[math]\hat{x}, \hat{y}, \hat{z}[/math] are clearly superior in this context yes.

Or just vector notation, it's not like the plus means anything here.

>Or just vector notation
Sure, but,

>it's not like the plus means anything here.
It absolutely does, and maybe the fact that you think this is a good argument for forcing students to use it occasionally.

Oh, it's to distinguish vectors from points...

No, the plus doesn't mean anything, you could just as well replace it with a "&" or "," or anything else you like to indicate another direction.

It does though...the way the product is defined makes the + the addition member of the ring.

Addition of three-vectors is meaningful, the addition of the three unit vectors is meaningless.
Which product?

>Addition of three-vectors is meaningful, the addition of the three unit vectors is meaningless.
These unit vectors ARE three-vectors. Please stop embarrassing yourself. Do you think (1, 0, 0) is equivalent to (1)?

Yes, but they don't interact, it's a waste of space to add them, just write them up as vectors.

> Which product?
quaternion product :^)
> Addition of three-vectors is meaningful
> the addition of the three unit vectors is meaningless.
unit vectors are vectors.

It also makes scaling, dot, and cross products look like distribution, instead of using linearity and bilinearity properties

7i+j+ak conveys no more information than (7,1,a) so the + is essentially meaningless.

They are not quaternions, they are three vectors, i=(1,0,0), j=(0,1,0), k=(0,0,1). I'm well aware that the notation makes sense with quaternions but it is absurd with three-vectors.

> but they don't interact
not in this context, but if you're using two different bases it's an incredibly useful notation

3 vectors ARE quaternions.
That's what Gibbs vs Hamilton's followers fought about.

You ignored my distribution looks cleaner comment

>let's loop back to
See , it has nothing to do with information, students should be exposed to different representations which may help you recognize properties they might have overlooked. As far as I'm concerned, we should do linear algebra with bra-ket, but I'm pretty sure there's still benefit to learning linear algebra syntax.

You've got the normal at that point right? The dot of that with the tangent is zero. Find a that makes the dot product zero. I think your normal should have a k component.
P.S. use exact values, those decimals are horrendous

Yeah, I'm more into math than statistics so I didn't look it up.

Sure, but if the notation is gibberish there probably isn't much to be learned from it.

Surely you could just take the length of the gradient at (7,1)?

The smart-ass answer to (a) is 0. The ass-free answer is the gradient of f at (5,2) or some positive scalar multiple thereof.

(b) The point on the surface is (5,2,f(5,2)). Figure that out.

Consider the line L(t) = (5+7 t,2+t) in the plane. This runs through (5,2) when t=0, and heads in the direction (7,1). Then the curve on the surface about this line is C(t) = (5+7 t, 2+ t, f(5+7 t, 2+t) ). Find C'(t) and evaluate that at t=0. This will give you the value for a.

In other words, if g(t)=f(5+7 t , 2+t), then
a=g'(0).

In other other words, "a" is equal to the directional derivative of f at (5,2) in the direction (7,1). It's the inner product of the gradient there and the vector (7,1).

You don't understand vectors as well as you think.

This.

Incidentally each notation gives a different vector space but they are isomorphic.

In one space your basis elements are (1,0,0), (0,1,0), (0,0,1) and in the other they are i, j, k.

The best part about this thread is that no one is even interested in OPs question.

what does statistics have to do with anything?
You don't know the distributive property?
> if the notation is gibberish
for you

>Smart-ass answer is a=0
Which is correct because it is the tangent line for at z=f(5,2) level curve? Nice haha.
Turns out the answer was just the dot product between the gradient of z and the stupid (7,1) vector.
I was so confused because I thought (7,1,a) was centered on the origin, in some tangent plane. I was literally getting suicidal before realizing that the vector was sprouting from the (5,2,f(5,2)) point.
I did that the moment I translated that i+7j+ak really meant (1,7,a) from the (5,2,f(5,2)) point.

I reckon I know them plenty for the purposes of this discussion.
You didn't call it the distributive property though, you called it distribution, and this I only associate with statistics and Sobolev spaces.
I understand the notation but it is cumbersome for no apparent reason.