$\begingroup$This was the question (or rather to prove) given by our professor: $|(a.c)| \leq |a||c|$ where $a$ and $c$ are vectors in $\mathbb{R}^n$. My friend and I solved it in two different ways and now we don't know who is correct. $a=\{x,y,z\}$ and $c=\{d,e,f\}$ to find $|\{x,y,x\}.\{d,e,f\}|$.
My approach: $\sqrt{(xd)^2 + (ye)^2 + (zf)^2}$
My friend's: $\sqrt{(xd+ye+zf)^2}$
Which one is correct? Thank You! (Bear with me if i am wrong with brackets... 'a' and 'c' are vectors.)
$\endgroup$41 Answer
$\begingroup$Using $$\displaystyle \vec{a}\cdot \vec{c}=|\vec{a}||\vec{c}|\cos \theta \leq |\vec{a}||\vec{c}|$$
Now Let $$\vec{a} = <x,y,z>$$ and $$\vec{c}=<d,e,f>$$
So Using above formula $$(\vec{a}\cdot \vec{c})\leq |\vec{a}||\vec{c}|\Rightarrow (\vec{a}\cdot \vec{c})^2\leq |\vec{a}|^2|\vec{c}|^2$$
So we get $$(xd+ye+zf)^2\leq (x^2+y^2+z^2)(d^2+e^2+f^2)$$
So $$(xd+ye+zf)\leq \sqrt{(x^2+y^2+z^2)(d^2+e^2+f^2)}$$
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