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