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10.3: Properties

  • Page ID
    92176
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    Commutativity

    Let's look at a few properties of the dot product. First of all, the dot product is commutative:

    \[\mathbf{A} \cdot \mathbf{B}=\mathbf{B} \cdot \mathbf{A}\]

    The proof of this property should be obvious from Eqs. (7.1.1) and (7.2.6). This isn't a trivial property; in fact, the other two types of vector multiplication are non-commutative.

    Projections

    The dot product is defined as it is because it gives the projection of one vector onto the direction of another. For example, dotting a vector \(\mathbf{A}\) with any of the cartesian unit vectors gives the projection of \(\mathbf{A}\) in that direction:

    \[
    \begin{align}
    \mathbf{A} \cdot \mathbf{i} & =A_{x} \\[6pt]
    \mathbf{A} \cdot \mathbf{j} & =A_{y} \\[6pt]
    \mathbf{A} \cdot \mathbf{k} & =A_{z}
    \end{align}
    \]

    In general, the projection of vector \(\mathbf{A}\) in the direction of unit vector \(\hat{\mathbf{u}}\) is \(\mathbf{A} \cdot \hat{\mathbf{u}}\).

    Magnitude

    From Eq. (7.2.6), it follows that \(\mathbf{A} \cdot \mathbf{A}=A_{x}^{2}+A_{y}^{2}+A_{z}^{2}=A^{2}\); so the magnitude of a vector \(\mathbf{A}\) is given in terms of the dot product by

    \[
    \begin{align}
    A^{2} & =\mathbf{A} \cdot \mathbf{A} \\[6pt]
    A & =\sqrt{\mathbf{A} \cdot \mathbf{A}}
    \end{align}
    \]

    Angle between Two Vectors

    The dot product is also useful for computing the separation angle between two vectors. From Eq. (7.1.1),

    \[\cos \theta=\frac{\mathbf{A} \cdot \mathbf{B}}{A B} \]

    Example \(\PageIndex{1}\)

    We wish to find the angle between the two vectors \(\mathbf{A}=3 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k}\) and \(\mathbf{B}=\mathbf{i}-5 \mathbf{j}+2 \mathbf{k}\).

    Solution

    We first find the dot product of the two vectors:

    \[\mathbf{A} \cdot \mathbf{B}=(3)(1)+(4)(-5)+(-2)(2)=-21 \notag \]

    The magnitudes of the two vectors are

    \[
    \begin{align}
    & A=\sqrt{3^{2}+4^{2}+(-2)^{2}}=\sqrt{29} \notag \\[6pt]
    & B=\sqrt{1^{2}+(-5)^{2}+2^{2}}=\sqrt{30} \notag
    \end{align} \notag
    \]

    Therefore

    \[\cos \theta=\frac{-21}{\sqrt{29} \sqrt{30}}=-0.711967 \notag \]

    and so the angle between \(\mathbf{A}\) and \(\mathbf{B}\) is

    \[\theta=135.40^{\circ} \notag \]

    You do not need to worry about getting the angle \(\theta\) in the correct quadrant, because \(\theta\) will necessarily always be between \(0^{\circ}\) and \(180^{\circ}\), and the inverse cosine function will always return its result in this range.

    Orthogonality

    Another useful property of the dot product is: if two vectors are orthogonal, then their dot product is zero. For example, for the cartesian unit vectors:

    \[\mathbf{i} \cdot \mathbf{j}=\mathbf{j} \cdot \mathbf{k}=\mathbf{i} \cdot \mathbf{k}=0\]

    The converse is also true: if the dot product is zero, then the two vectors are orthogonal.

    The cartesian unit vectors \(\mathbf{i}, \mathbf{j}\), and \(\mathbf{k}\) are orthonormal, so that

    \[\mathbf{i} \cdot \mathbf{i}=\mathbf{j} \cdot \mathbf{j}=\mathbf{k} \cdot \mathbf{k}=1\]

    Derivative

    The derivative of the dot product is similar to the familiar product rule for scalars:

    \[\frac{d(\mathbf{A} \cdot \mathbf{B})}{d t}=\mathbf{A} \cdot \frac{d \mathbf{B}}{d t}+\frac{d \mathbf{A}}{d t} \cdot \mathbf{B}\]


    10.3: Properties is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.