Question

Determine as componentes e o módulo de $\vec r = \vec a - \vec b + \vec c$, onde $\vec a = 5 \hat i + 4 \hat j - 6 \hat k$, $\vec b = -2 \hat i + 2 \hat j + 3 \hat k$ e $\vec c = 4 \hat i + 3 \hat j + 2 \hat k$. Calcule o ângulo entre $\vec r$ e o eixo z positivo.

Answer 1

cálculo do vetor $\sf{\vec{r}}$ :

$\sf{\vec{r}=(5\hat{i}+4\hat{j}-6\hat{k})-(-2\hat{i}+2\hat{j}+3\hat{k})+(4\hat{i}+3\hat{j}+2\hat{k})}\\\sf{5\hat{i}+2\hat{i}+4\hat{i}+4\hat{j}-2\hat{j}+3\hat{j}-6\hat{k}-3\hat{k}+2\hat{k}}\\\boxed{\boxed{\sf{\vec{r}=11\hat{i}+5\hat{j}-7\hat{k}}}}$

Façamos o produto escalar entre $\sf{\vec{r}~e~\hat{k}}$ :

$\sf{\bf{r\cdot k}=r\cdot1\cdot cos(\theta)}\\\sf{cos(\theta)=\dfrac{\bf{r\cdot k}}{r}}$

$\sf{\bf{r\cdot k}=(11\hat{i}+5\hat{j}-7\hat{k})\cdot\hat{k}=0+0-11=-7}$

calculando r temos:

$\sf{r=\sqrt{11^2+5^2+(-7)^2}}\\\sf{r=\sqrt{121+25+49}=\sqrt{195}\approx~13,96}$

Daí

$\sf{cos(\theta)=\dfrac{-7}{13,96}}\\\sf{cos(\theta)=-0,50}\\\sf{\theta=arccos(-0,50)}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{\theta\approx~120^{\circ}}}}}}$