Question

Let u and v=24 Check if ||u||+||v|| = ||u+v||

Answer 1

well, I'm assuming v = 24, is written in a+bi form, that is v = 24 + 0i, or <24, 0 >

Answer 2

Im not sure about that... the problem is written just like I did

Answer 3

I'd think is 2 vectors, one written in an ordered pair, and the other as a complex form, either way a + bi = <a, b>

Answer 4

as far as the ||u|| that, is the "magnitude" of the vector is really just \(\huge ||v|| = \sqrt{a^2+b^2}\)

Answer 5

and vector addition is just <a, b> + <c, d> => <a+c, b+d>

Answer 6

so, ||<a, b>|| + ||<24, 0>|| => \(\sqrt{a^2+b^2} + \sqrt{24}\)

Answer 7

||u+v|| that is => || <a,b> + <24, 0> ||
=> || <a+24, b> ||
=> \(\sqrt{(a+24)^2+b^2}\\
\implies \sqrt{a^2+48a+24^2+b^2}\)

Answer 8

so, I gather the answer, they're \(\ne\)

Answer 9

ohh and
so, ||<a, b>|| + ||<24, 0>|| => \(\sqrt{a^2+b^2} + \sqrt{24^2}\)

Answer 10

Let v be <c,d>\(\sqrt{a^2+b^2}+24 \neq \sqrt{(a+c)^2+(b+d)^2}=\sqrt{a^2+2ac+c^2+b^2+2bd+d^2}=\)
\(=\sqrt{a^2+2(ac+bd)+b^2 +24^2}\)

Answer 11

@Pradius

Answer 12

gimme a sec