Q1.5 Multiplying vectors: the dot (or scalar) product (S)

Write down the definition of the dot product of two vectors.

The three vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ form a right-angled triangle in which B=2 and C=1 (in arbitrary units).

  1. Determine the values of

    $\vec{A} \cdot \vec{B}$ and $\vec{A} \cdot \vec{C}$

  2. Compare $\vec{A} \cdot \vec{B}$ with $\vec{B} \cdot \vec{A}$ and comment.
  3. Interpret your results geometrically.
Graphic - No title - ABCtriangle
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Hint

Does the order of the vectors matter in the dot product?
 

Solution

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Solution

The definition of the dot product of two vectors –let’s call them $\vec{D}$ and $\vec{E}$ here– is (Key Point 0.3)
\[ \vec{D}\cdot\vec{E} = DE \cos \theta \]
where θ is the angle between the two vectors.
  1. It follows that in this case

    \[ \vec{A}\cdot\vec{B} = AB \cos {\theta_{AB}} = AB \times\frac{B}{A} =B^2 = 4 \]
    while
    \[ \vec{A}\cdot\vec{C} = AC \cos {\theta_{AC}} = AC \times[-\frac{C}{A}] =-C^2 =-1 \]

    The minus sign comes from the fact that the angle between $\vec{A}$ (you need to extend it) and $\vec{C}$ is greater than 90.

  2. From the definition it follows that
    \[ \vec{A}\cdot\vec{B} = \mbox{\rm magnitude of A} \times \mbox{\rm magnitude of B} \times \mbox{\rm cos of angle between them} \]
    is the same as $\vec{A}\cdot\vec{B}$. The order of the vectors in a scalar product doesn’t matter.
  3. Finally the geometric interpretation of $\vec{D}\cdot \vec{E}$ is that it gives the projection of one vector (say the first) onto the second, times the magnitude of the second. In this case the projection of $\vec{A}$ on $\vec{B}$ is just B so that $\vec{A}\cdot\vec{B}$ is indeed B2. A similar argument recovers our result for $\vec{A}\cdot\vec{C}$