Question

Vector question below!

Answer 1

Under what conditions is ||u||+||v||=||u+v||? Wherein u and v are vectors.

Answer 2

So when u=<0,0,0> and v=<0,0,0>... I'm not sure about what other possibilities are there.

Answer 3

No vectors given.

Answer 4

by definition, what is |u+v| ? or |u+v| =??

Answer 5

ok, I ask you another thing. |a| = <a, a> , right?

Answer 6

I will start helping when you reply.

Answer 7

@Loser66 Sorry for the late reply! ||u+v|| is the magnitude of the resultant vector from u+v. There wasn't much information given, just this one line. Thanks for your reply!

Answer 8

do you need the answer? @mortonsalt

Answer 9

@openstudygirl2 I do, but I'd like to know the method more, tbh. It's a tricky one :) Thanks!

Answer 10

uhh well do you wanna try my way quickly ??

Answer 11

......

Answer 12

ok she left

Answer 13

This of vectors as distance from origin



When is the distance from a to b plus the distance from b to c going to be the same as the distance from a to c?

Answer 14

Think of vectors as distance from origin*

Answer 15

^that diagram should help you visualize it...

Answer 16

\(||u+v|| =||u ||+||v||\) if and only if \(||u+v||^2=(||u||+||v||)^2\)

LHS
\(||a|| =\sqrt{<a,a>}\\||a||^2=<a,a>\)

Hence \(||u+v||^2=<u+v,u+v>\\ = <u,u> +2<u,v>+<v,v>\\=||u||^2+2<u,v>+||v||^2\)

RHS
\((||u||+||v||)^2=||u||^2+2||u||||v||+||v||^2\)

Answer 17

then LHS = RHS iff

\(<u,v> =||u||||v||\)

But you know that \(<u,v> = ||u||||v||\) iff \(\vec u\ // \vec v\)

why? because \(cos (\theta)=\dfrac{<u,v>}{||u||v||}\)

where \(\theta \) is the angle between u and v

And we want \(cos (\theta)=1\) to get the equation above. Hence u // v is the required condition.

Answer 18

@openstudygirl2 @zzr0ck3r @agent0smith @Loser66 Thanks so much for your responses! I apologise for not getting back to you guys ASAP. I think I understand it now, thanks!