$10 to the first person who answers ii and iii

$10 to the first person who answers ii and iii

Attached: Snip20191118_1.png (1252x642, 182K)

Why don't you do your own homework instead? You're only cheating yourself out of an education.

i learn from examples bro

i do agree completely with what u say though, people do cheat themselves

the answer is 7 and the other one is yes

Bro you will never graduate high school if you can't figure this out

this is second year maths degree lol

Well if you were paying attention in class, there have probably been quite a few examples of similar problems.

>American education

no, the lecturer wasn't in for this lecture so he skipped this part, and the notes do not contain anything like this

No that would be the British

Alright. I'm not going to do your homework for you, but I can at least give you a place to start.
ii) wants you to use scalar products to show something. So do some scalar products.
A dot B = |A|*|B|*cos(theta)
They've given you 3 vectors. n, SP, HP
Try computing n dot SP, n dot HP, and simplifying the results to see if they give you the result you're looking for.

wait why would u do the dot product of two separate times

ii iii

also sp and hp are not vectors they are coordinates

Because they want you to show that both theta and phi equal ab^2/|n|
theta is the angle between n and SP. phi is the angle between n and HP, so you've got to compute both
False. S, H, and P are points. SP and HP are vectors

i thought it was the dot product of;
((x1+ae),y1) . (b^2x1,a^2y2)

Ohh sorry, I just don't understand how they simplify into that part. That's my only issue with part ii

Let F and U be my response to that question, and take this whole thing and shove it up your ass.

It looks like it's going to be a bit of algebra.
A dot B = Ax*Bx+Ay*By
You know the x and y components of n, SP, and HP
So plug into that formula and massage the result until it looks like the result you're supposed to get

What kind of class is this and what can you use? I'm using notation to denote inner (cross) product between vectors x and y.

For (i), just pick either positive or negative y values to get a proper function for y(x), derivate to get y'(x), think of y'(x) as a vector V of arbitrary length and use orthogonality to find the normal vector T=(t1,t2) since =0 (Solve for T) iff those two are orthogonal. Then once you have the orthogonal vector T, it's just a matter of finding a scalar C such that P+CT=n.

Whatever the steps for (ii) are, just turn every line into a vector and use the fact that ||=|x||y|cos(a), where a is the angle between the two vectors.You'll get there automatically if you consider that, unless you're a huge retard.

For (iii) use basic geometry as well as the facts about orthogonality and the norm of the inner product you used in (i) and (ii).

>Second year math degree
What shithole school is this? I'm a geophysicist and did the same math that math majors do their first two years, and this was basic first year stuff from real analysis.

pretty much this
though you presumably meant dot product instead of cross product

Whoops hurr durr, yes.

But every time I expand this i get fucked figures

Attached: 08E73A91-23B8-4C49-9F89-C498118A0CBE.jpg (640x480, 37K)

its an economic math module

did u use ((x1+ae),y) as your point

Nigger what are you even doing there, I legitimately have no clue.

Just listen to me. Do y'(x), I don't see you doing that anywhere. Once you do every vector V is gonna be (1, y(x)) for a given x. Then find T such that =0. Protip: T = (1, -1/y'(x)), do it symbolically and it should work no matter what y(x) is. There, you have the normal vector T. The derivative looks messy at first glance but it's a simple application of the chain rule concerning a square root and x squared, and then all you need to do is invert it and slap a negative on the front to get the second component of the normal vectorz the first component being 1 as we said last sentence. After that write an equality term-to-term for P+CT=n, and solve for C; if it exists, your proof is done. Is there really a part of this you don't understand?

i still dont undertsand the simplification

how does that include the scalar product lol

I haven't worked out all the math. This is your homework, not mine.
But
n = (b^2x1, a^2y1)
SP = P-S = (x1+ae, y1)
Then dot them and simplify
So yes, one of the vectors is (x1+ae,y1)

yeah thats the thing i can never simplify from that point. i can neevr geet the root at the bottom to cancel out

and you cant rationalise the denominator as then u would have to solve for |n|

ii - No
iii True

those are answers, they are not correct but answers

Nailed it mate

welll it was a good attempt guys

still, have no clue on how to simplify