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Physics LibreTexts

25.7: Sample problems and solutions

  • Page ID
    19569
  • 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: \[\begin{aligned} \vec d = \begin{pmatrix} 4\\ 5\\ 6\\ \end{pmatrix} - \begin{pmatrix} 1\\ 2\\ 3\\ \end{pmatrix}=\begin{pmatrix} 3\\ 3\\ 3\\ \end{pmatrix}\end{aligned}\]

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

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