5. Vectors

d. Dot Product

2. Properties

We here list the algebraic properties of the dot product and its relation to magnitude, vector addition and scalar multiplication.

In the following, the zero vector, denoted by \(\vec 0\), is the vector whose components are all zero and the negative of a vector \(\vec v\), denoted by \(-\vec v\), is the vector whose components are the negatives of those of \(\vec v\).

First, there is one property involving just the dot product.

Let \(\vec u\) and \(\vec v\) be arbitrary vectors. Then,

The Dot Product is Commutative: \(\vec u\cdot\vec v=\vec v\cdot\vec u\)

\[ \vec u\cdot\vec v=u_1v_1+u_2v_2 =v_1u_1+v_2u_2=\vec v\cdot\vec u \]

Second, there is one relation between the dot product and magnitude.

Let \(\vec v\) be an arbitrary vector. Then,

Square Property: \(\vec v\cdot\vec v=|\vec v|^2\)

\[ \vec v\cdot\vec v =v_1v_1+v_2v_2 =|\vec v|^2 \]

Third, there are two properties relating the dot product and vector addition.

Let \(\vec u\), \(\vec v\) and \(\vec w\) be arbitrary vectors. Then,

The Dot Product Distributes over Vector Addition: \(\vec u\cdot(\vec v+\vec w) =\vec u\cdot\vec v+\vec u\cdot\vec w\)

\[\begin{aligned} \vec u\cdot(\vec v+\vec w) &=\left\langle u_1,u_2\right\rangle\cdot \left\langle v_1+w_1,v_2+w_2\right\rangle \\ &=u_1(v_1+w_1)+u_2(v_2+w_2) \\ &=u_1v_1+u_2v_2+u_1w_1+u_2w_2 \\ &=\vec u\cdot\vec v+\vec u\cdot\vec w \end{aligned}\]

The Dot Product of \(\vec 0\) and any vector is \(0\): \(\vec 0\cdot\vec v=0\)

\[ \vec 0\cdot\vec v =0v_1+0v_2=0 \]

Since the dot product is commutative, it follows that in these two formulas the dot products can be done in the opposite order: \[ (\vec v+\vec w)\cdot\vec u =\vec v\cdot\vec u+\vec w\cdot\vec u \] \[ \vec v\cdot\vec 0=0 \]

Finally, we have two relations between the dot product and scalar multiplication.

Let \(\vec u\) and \(\vec v\) be arbitrary vectors and \(a\) be an arbitrary scalar. Then,

The Dot Product Associates with Scalar Multiplication: \((a\vec u)\cdot\vec v=a(\vec u\cdot\vec v)\)

\[\begin{aligned} (a\vec u)\cdot\vec v &=\left\langle au_1,au_2\right\rangle\cdot \left\langle v_1,v_2\right\rangle \\ &=au_1v_1+au_2v_2 \\ &=a(u_1v_1+u_2v_2) =a(\vec u\cdot\vec v) \end{aligned}\]

The Dot Product Commutes with Scalar Multiplication: \((a\vec u)\cdot\vec v=\vec u\cdot(a\vec v)\)

Use components. \[\begin{aligned} (a\vec u)\cdot\vec v &=\left\langle au_1,au_2\right\rangle\cdot \left\langle v_1,v_2\right\rangle \\ &=au_1v_1+au_2v_2 \\ &=u_1av_1+u_2av_2 \\ &=\left\langle u_1,u_2\right\rangle\cdot \left\langle av_1,av_2\right\rangle \\ &=\vec u\cdot(a\vec v) \end{aligned}\]

Use previous properties. \[\begin{aligned} (a\vec u)\cdot\vec v &=a(\vec u\cdot\vec v) &\quad&\text{Associative}\\ &=a(\vec v\cdot\vec u) &\quad&\text{Commutative}\\ &=(a\vec v)\cdot\vec u &\quad&\text{Associative}\\ &=\vec u\cdot(a\vec v) &\quad&\text{Commutative}\\ \end{aligned}\]

the dot product is commutative using the vectors \(\vec u=\left\langle 2,11\right\rangle\) and \(\vec v=\left\langle -3,1\right\rangle\).

Verify means to check that a property works for specific vectors. It is not a proof!

We compute both sides of the equation: \(\vec u\cdot\vec v=\vec v\cdot\vec u\). \[\begin{aligned} \vec u\cdot\vec v &=\left\langle 2,11\right\rangle\cdot\left\langle -3,1\right\rangle =(2)(-3)+(11)(1)=5 \\ \vec v\cdot\vec u &=\left\langle -3,1\right\rangle\cdot\left\langle 2,11\right\rangle =(-3)(2)+(1)(11)=5 \end{aligned}\] Since both sides give the same answer, the commutative property is verified (not proved).

Verify the dot product is distributive using the vectors \(\vec u=\left\langle 2,11\right\rangle\), \(\vec v=\left\langle -3,1\right\rangle\), and \(\vec w=\left\langle 1,2\right\rangle\).

Compute the left and right sides of the identity: \[ \vec u\cdot(\vec v+\vec w)=\vec u\cdot\vec v+\vec u\cdot\vec w \]

We first compute the left side: \[\begin{aligned} \vec u\cdot(\vec v+\vec w) &=\left\langle 2,11\right\rangle\cdot \big(\left\langle -3,1\right\rangle+\left\langle 1,2\right\rangle\big) \\ &=\left\langle 2,11\right\rangle\cdot\left\langle -2,3\right\rangle \\ &=-4+33 =29 \end{aligned}\] Second we compute the right said: \[\begin{aligned} \vec u\cdot\vec v+\vec u\cdot\vec w &=\left\langle 2,11\right\rangle\cdot\left\langle -3,1\right\rangle +\left\langle 2,11\right\rangle\cdot\left\langle 1,2\right\rangle \\ &=(-6+11)+(2+22) \\ &=5+24 =29 \end{aligned}\] Since both sides give the same answer, the distributive property is verified.

5. Vectors

d. Dot Product

2. Properties

Heading