# 25.7: Sample problems and solutions

- Page ID
- 19569

Exercise \(\PageIndex{1}\)

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

**Answer**-
a. The displacement vector is given by: \[\begin{aligned} \vec d = \begin{pmatrix} 4\\ 5\\ 6\\ \end{pmatrix} - \begin{pmatrix} 1\\ 2\\ 3\\ \end{pmatrix}=\begin{pmatrix} 3\\ 3\\ 3\\ \end{pmatrix}\end{aligned}\]

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