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25.7: Sample problems and solutions

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    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:

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

    This page titled 25.7: Sample problems and solutions is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Ryan D. Martin, Emma Neary, Joshua Rinaldo, and Olivia Woodman via source content that was edited to the style and standards of the LibreTexts platform.