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Here"s a simple proof:
$|vecxcdotvecy| leq |vecx||vecy| $
Substitute $|vecxcdotvecy| = |vecx||vecy|cos heta$
$| |vecx||vecy|cos heta |leq |vecx||vecy| $
Divide both sides by $|vecx||vecy|$
$ | cos heta| leq 1$
-Hey, I was looking for a "more serious" proof!
Then here you are!
Here"s another simple proof:
This is projecting a vector to lớn another one (Click the gif if it doesn"t load):
You drag its end in a line that is perpendicular to the other vector. Then multiply the length of the new vector with the old vector.
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Do you know what the multiplication is equal to? The dot sản phẩm of the vectors
When you project that vector, its norm (length) becomes lower - or stays the same if one of them is a scalar multiple of the other one.
^^ That was the proof. Think about it.
Source: $3$Blue$1$Brown
Wait, I look for a "really serious" proof!
Here you are.
Another proof:
Let $p(t)=||tvecy-vecx||^2$
As there"s an absolute value, it must be equal khổng lồ or bigger than $0$.
$p(t)=||tvecy-vecx||^2geq 0$
$p(t)=(tvecy-vecx)(tvecy-vecx)geq 0$
$p(t)=t^2(vecycdot vecy)-2t(vecxcdotvecy)+vecxcdot vecxgeq0$
Let"s substitute some things.
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$p(t)=t^2underbrace(vecycdot vecy)_colorbluelarge a+tunderbrace(-2vecxcdotvecy)_colorredlarge b+underbrace(vecxcdot vecx)_colorgreenlarge cgeq0$
$p(t)=colorblueat^2+colorredbt+colorgreencgeq0$
Its minimum value must be $large frac-colorredb2colorbluea$
Substituting $large t= frac-colorredb2colorbluea$
$p(frac-colorredb2colorbluea)=colorbluea(frac-colorredb2colorbluea)^2+colorredb(frac-colorredb2colorbluea)+colorgreencgeq0$
$p(frac-colorredb2colorbluea)=colorbluea(fraccolorredb^24colorbluea^2)+colorredb(frac-colorredb2colorbluea)+colorgreencgeq0$
$p(frac-colorredb2colorbluea)=fraccolorredb^24colorbluea+frac-colorredb^22colorbluea+colorgreencgeq0$
Forget the $large p(t)$ function side (LHS)
$fraccolorredb^24colorbluea+frac-colorredb^22colorbluea+colorgreencgeq0$
Multiply by $large 4colorbluea$
$colorredb^2-2colorredb^2+4colorblueacolorgreencgeq0$
$-colorredb^2+4colorblueacolorgreencgeq0$
$4colorblueacolorgreencgeq colorredb^2$
De-substitute
$p(t)=t^2underbrace(vecycdot vecy)_colorbluelarge a+tunderbrace(-2vecxcdotvecy)_colorredlarge b+underbrace(vecxcdot vecx)_colorgreenlarge cgeq0$
$4colorblue(vecycdot vecy)colorgreen(vecxcdot vecx)geq colorred(-2vecxcdotvecy)^2$
Using the identity $large vecvcdotvecv=||vecv||^2$
$4colorbluecolorgreengeq colorred(-2vecxcdotvecy)^2$
Using the identity $(f(x))^2=(|f(x)|)^2$ (where $f(x)inkumeR$)
$4colorbluecolorgreengeq colorred-2vecxcdotvecy^2$
As the both sides are not negative, you can square root both sides.
$2colorbluecolorgreengeq colorred$
$2colorbluecolorgreengeq colorredvecxcdotvecy$
$largecolorbluecolorgreengeq colorred$