[Boards: 3 / a / aco / adv / an / asp / b / biz / c / cgl / ck / cm / co / d / diy / e / fa / fit / g / gd / gif / h / hc / his / hm / hr / i / ic / int / jp / k / lgbt / lit / m / mlp / mu / n / news / o / out / p / po / pol / qa / qst / r / r9k / s / s4s / sci / soc / sp / t / tg / toy / trash / trv / tv / u / v / vg / vip /vp / vr / w / wg / wsg / wsr / x / y ] [Search | Home]
a little help please....
If images are not shown try to refresh the page. If you like this website, please disable any AdBlock software!

You are currently reading a thread in /sci/ - Science & Math

File: ggg.jpg (14 KB, 342x281) Image search: [iqdb] [SauceNao] [Google]
14 KB, 342x281
prove the scalar product of AC.DF=0
knowing that AB=a
>>
DF = DH+HG+GF
Then expand the dot product
>>
>>7785252
please continue in more detail.thx
>>
>>7785252
so it's possible to expand the dot product the same way as any algebric product ?
and the answer becomes;
>>
>>7785296
>so it's possible to expand the dot product the same way as any algebric product ?
Yes.
Better make this step more detailed.
DC.DH = 0
DC.HG = 2|a|
DC.GF = 0
>>
>>7785304
ok thx
>>
>>7785304
DC.HG = |a|^2
>>
There are a few ways to solve this problem, some more intuitive than others.

AC = a(-i + j)
DF = a(i + j + k)

AC.DF = -a^2 + a^2 + 0 = 0

An even easier way would be to evaluate it analytically. AC is inverse to DB, and because AC omits the extra dimension that DF posses, then we can determine the dot product between the two vectors to be 0 (with the angle between them being pi/2 radians).

>>7785260
Stop being an ungrateful dick and learn the material for yourself. You aren't going to benefit if people just tell you. Be lucky that people are willing to help you even when you're being a selfish pig.