25.7: Sample problems and solutions

Exercise $\PageIndex{1}$

1. What is the displacement vector from position $(1,2,3)$ to position $(4,5,6)$?
2. What angle does that displacement vector make with the $x$ axis?

Answer

a. The displacement vector is given by:

$$\begin{aligned} \vec d = \begin{pmatrix} 4\\[4pt] 5\\[4pt] 6\\[4pt] \end{pmatrix} - \begin{pmatrix} 1\\[4pt] 2\\[4pt] 3\\[4pt] \end{pmatrix}=\begin{pmatrix} 3\\[4pt] 3\\[4pt] 3\\[4pt] \end{pmatrix}\end{aligned}$$

b. We can find the angle that this vector makes with the $x$ axis by taking the scalar product of the displacement vector and the unit vector in the $x$ direction (1,0,0):

$$\begin{aligned} \hat x \cdot \vec d = (1)(3)+(0)(3)+(0)(3) = 3\end{aligned}$$

This is equal to the product of the magnitude of $\hat x$ and $\vec d$ multiplied by the cosine of the angle between them:

$$\begin{aligned} \hat x \cdot \vec d &= ||\hat x||||\vec d||\cos\theta = (1)(\sqrt{3^2+3^2+3^2})\cos\theta= \sqrt{27}\cos\theta\\[4pt] 3 &= \sqrt{27}\cos\theta\\[4pt] \therefore \cos\theta &= \frac{3}{\sqrt{27}} = \frac{1}{\sqrt{3}}\\[4pt] \theta&=54.7^{\circ}\end{aligned}$$