MATH 151 Chapter 1.2

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Dot Product

Definition

Also called scalar product because it returns a scalar number for an answer

$\mathbf {a} \cdot \mathbf {b} =\left|\mathbf {a} \right|\left|\mathbf {b} \right|\cos {\theta }$

Where θ is the angle between a and b

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Given $\mathbf {a} =\left\langle a_{1},\,a_{2}\right\rangle$ and $\mathbf {b} =\left\langle b_{1},\,b_{2}\right\rangle$,

$\mathbf {a} \cdot \mathbf {b} =a_{1}b_{1}+a_{2}b_{2}$

Example

A crate is hauled 8 meters up a ramp under a constant force of 20 Newtons applied at an angle of 25° to the ramp. Find the work done.

${\begin{aligned}W&=\mathbf {F} \cdot \mathbf {D} \\&=\left|\mathbf {F} \right|\left|\mathbf {D} \right|\cos {25^{\circ }}\\&=\left(20\right)\left(8\right)\cos {25^{\circ }}\\&\approx 145\ {\mbox{J}}\end{aligned}}$

Orthogonal (Perpendicular) Vectors

Two vectors a and b are orthogonal iff their dot product is 0

Orthogonal Complement

Given the nonzero vector $\mathbf {a} =\left\langle a_{1},\,a_{2}\right\rangle$, the orthogonal complement of a is the vector $\mathbf {a} ^{\perp }=\left\langle -a_{2},\,a_{1}\right\rangle$.

Projections

Projection of one vector a onto another vector b results in a component of a that is along b

2 types: scalar and vector projections:

• scalar projection function written as "comp_a b"
• vector projection function written as "proj_a b"

Read as "scalar/vector projection of b onto a"

think of the letter b in comp_a b or proj_a b as being "on top of" the a.

Scalar Projection (results in a scalar answer)

${\mbox{comp}}_{\mathbf {a} }\,\mathbf {b} ={\frac {\mathbf {a} \cdot \mathbf {b} }{\left|\mathbf {a} \right|}}$

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Vector Projection (results in a vector answer)

${\begin{aligned}{\mbox{proj}}_{\mathbf {a} }\,\mathbf {b} &={\frac {\mathbf {a} \cdot \mathbf {b} }{\left|\mathbf {a} \right|^{2}}}\mathbf {a} \\&=\left({\mbox{comp}}_{\mathbf {a} }\,\mathbf {b} \right)\times \mathbf {u_{a}} \qquad \left(\mathbf {u_{a}} \ {\mbox{is unit vector along}}\ \mathbf {a} \right)\end{aligned}}$

Angle Between Vectors

If we define $\mathbf {a} \cdot \mathbf {b} =\left|\mathbf {a} \right|\left|\mathbf {b} \right|\cos {\theta }$, then

$\theta =\arccos {\left({\frac {\mathbf {a} \cdot \mathbf {b} }{\left|\mathbf {a} \right|\left|\mathbf {b} \right|}}\right)}$