Question

dados u = (2,3,-2) e v = (1,4,6), determine:
a) o produto escalar entre esses vetores
b) o produto vetorial entre os vetores u e v ( nessa ordem)
c) a area do paralelograma cujos os lados soa os vetores u e v (nessa ordem)

Answer 1

$\vec{u}=(u_{1},~u_{2},~...~,~u_{n})\\\vec{v}=(v_{1},~v_{2},~...~,~v_{n})$

Produto escalar entre esses vetores:

$\boxed{\boxed{\vec{u}\cdot\vec{v}=\sum\limits_{i=1}^{n}u_{i}v_{i}}}$

Produto vetorial entre dois vetores do R³:

$\vec{u}=(u_{1},~u_{2},~u_{3})\\\vec{v}=(v_{1},~v_{2},~v_{3})\\\\\\\boxed{\boxed{\vec{u}~x~\vec{v}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\u_{1}&u_{2}&u_{3}\\v_{1}&v_{2}&v_{3}\end{array}\right|}}$

Onde:

$\vec{i}=(1,0,0)\\\vec{j}=(0,1,0)\\\vec{k}=(0,0,1)$

Área do paralelogramo formado por dois vetores no R³:

$\boxed{\boxed{A=||\vec{u}~x~\vec{v}||}}$
________________________________

$\vec{u}=(2,~3,-2)\\\vec{v}=(1,~4,~6)$

a)

$\vec{u}\cdot\vec{v}=\sum\limits_{i=1}^{3}u_{i}v_{i}\\\\\\\vec{u}\cdot\vec{v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}\\\\\vec{u}\cdot\vec{v}=2\cdot1+3\cdot4-2\cdot6\\\\\vec{u}\cdot\vec{v}=2+12-12\\\\\boxed{\boxed{\vec{u}\cdot\vec{v}=2}}$

b)

$\vec{u}~x~\vec{v}=det\left[\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\u_{1}&u_{2}&u_{3}\\v_{1}&v_{2}&v_{3}\end{array}\right]\\\\\\\vec{u}~x~\vec{v}=det\left[\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\2&3&-2\\1&4&6\end{array}\right]\\\\\\\vec{u}~x~\vec{v}=(\vec{i}\cdot3\cdot6)+(\vec{j}\cdot[-2]\cdot1)+(2\cdot4\cdot\vec{k})-(\vec{k}\cdot3\cdot1)-(\vec{j}\cdot2\cdot6)-([-2]\cdot4\cdot\vec{i})\\\\\\\vec{u}~x~\vec{v}=18\vec{i}-2\vec{j}+8\vec{k}-3\vec{k}-12\vec{j}+8\vec{i}\\\\\\\vec{u}~x~\vec{v}=26\vec{i}-14\vec{j}+5\vec{k}$

$\boxed{\boxed{\vec{u}~x~\vec{v}=(26,-14,~5)}}$

c)

$A=||\vec{u}~x~\vec{v}||\\\\A=||(26,-14,~5)||\\\\A=\sqrt{26^{2}+(-14)^{2}+5^{2}}\\\\A=\sqrt{676+196+25}\\\\\boxed{\boxed{A=\sqrt{897}~(u.a)}}$