Using the Dot Product to Determine an Angle What is the angle θ between the lines AB and AC in Fig. 2.36? Strategy We know the coordinates of the points A, B, and C, so we can determine the components of the vector [latex]r_{AB}[/latex] from A to B and the vector [latex]r_{AC}[/latex] frorn A to C (Fig. a). Then we can use Eq. (2.24) to determine θ. [latex]\cos θ = \frac{U . V }{\left|U\right|\left|V\right|. } = \frac{U_{x} V_{x} + U_{y} V_{y} + U_{z} V_{z} }{\left|U\right|\left|V\right|} . [/latex] (2.24)

The vectors [latex]r_{AB}[/latex] and [latex]r_{AC}[/latex] are [latex]r_{AB} = (6 - 4)i + (1 - 3)j + (-2 - 2)k = 2i - 2j - 4k (m) , [/latex] [latex]r_{AC} = (8 - 4)i + (8 - 3)j + (4 - 2)k = 4i + 5j + 2k (m) .[/latex] Their magnitudes are [latex] \left|r_{AB}\right| = \sqrt{(2)^{2} + (- 2)^{2} + (- 4)^{2} } = 4.90 m , [/latex] [latex] \left|r_{AC}\right| = \sqrt{(4)^{2} + (5)^{2} + (2)^{2} } = 6.71 m . [/latex] The dot product of [latex]r_{AB}[/latex] and [latex]r_{AB}[/latex] is [latex]r_{AB} \cdot r_{AC} = (2)(4) + (- 2) (5) + (- 4)(2) = - 10 m^{2} [/latex] Therefore [latex]\cos θ = \frac{r_{AB} \cdot r_{AC} }{\left|r_{AB} \right|\left|r_{AC} \right| } = \frac{- 10}{(4.90)(6.71)} = - 0.304 . [/latex] The angle θ = arccos ( - 0.304) = 107.7°.

Question 2.14: Using the Dot Product to Determine an Angle What is the angl...

Engineering Mechanics: Statics

Using the Dot Product to Determine an Angle

What is the angle θ between the lines AB and AC in Fig. 2.36?

Strategy
We know the coordinates of the points A, B, and C, so we can determine the components of the vector $r_{AB}$ from A to B and the vector $r_{AC}$ frorn A to C (Fig. a). Then we can use Eq. (2.24) to determine θ.

$\cos θ = \frac{U . V }{\left|U\right|\left|V\right|. } = \frac{U_{x} V_{x} + U_{y} V_{y} + U_{z} V_{z} }{\left|U\right|\left|V\right|} .$ (2.24)

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