 maths > vector-algebra

Vector Dot Product : Component Form

what you'll learn...

Overview

Vector Dot Product : Component Form

»  sum of product of individual components
→  $\stackrel{\to }{p}\cdot \stackrel{\to }{q}= {p}_{x}{q}_{x}+{p}_{y}{q}_{y}+{p}_{z}{q}_{z}$

simple in component form

Given the two vectors
$\stackrel{\to }{p}={p}_{x}i+{p}_{y}j+{p}_{z}k$$\vec{p} = {p}_{x} i + {p}_{y} j + {p}_{z} k$

$\stackrel{\to }{q}={q}_{x}i+{q}_{y}j+{q}_{z}k$$\vec{q} = {q}_{x} i + {q}_{y} j + {q}_{z} k$ The vector dot product is defined as $|p||q|\mathrm{cos}\theta$$| p | | q | \cos \theta$

How will one compute the angle $\theta$$\theta$ from the given component forms of vectors? Consider this as triangle in coordinate plane. The triangle is made of 3 sides having scalar quantities $p=|\stackrel{\to }{p}|$$p = | \vec{p} |$, $q=|\stackrel{\to }{q}|$$q = | \vec{q} |$ and $r=|\stackrel{\to }{q}-\stackrel{\to }{p}|$$r = | \vec{q} - \vec{p} |$. The cosine rule of triangle is applicable.
${r}^{2}={p}^{2}+{q}^{2}-2pq\mathrm{cos}\theta$${r}^{2} = {p}^{2} + {q}^{2} - 2 p q \cos \theta$

${r}^{2}={\left({q}_{x}-{p}_{x}\right)}^{2}+{\left({q}_{y}-{p}_{y}\right)}^{2}+{\left({q}_{z}-{p}_{z}\right)}^{2}$${r}^{2} = {\left({q}_{x} - {p}_{x}\right)}^{2} + {\left({q}_{y} - {p}_{y}\right)}^{2} + {\left({q}_{z} - {p}_{z}\right)}^{2}$,
${p}^{2}={p}_{x}^{2}+{p}_{y}^{2}+{p}_{z}^{2}$${p}^{2} = {p}_{x}^{2} + {p}_{y}^{2} + {p}_{z}^{2}$,
${q}^{2}={q}_{x}^{2}+{q}_{y}^{2}+{q}_{z}^{2}$${q}^{2} = {q}_{x}^{2} + {q}_{y}^{2} + {q}_{z}^{2}$,

With algebraic manipulations on this, we can derive that
$\mathrm{cos}\theta =\frac{{p}_{x}{q}_{x}+{p}_{y}{q}_{y}+{p}_{z}{q}_{z}}{|p||q|}$$\cos \theta = \frac{{p}_{x} {q}_{x} + {p}_{y} {q}_{y} + {p}_{z} {q}_{z}}{| p | | q |}$
Substituting the above in the vector dot product we get.
$\stackrel{\to }{p}\cdot \stackrel{\to }{q}$$\vec{p} \cdot \vec{q}$

That derives the component form of vector dot product as
$\stackrel{\to }{p}\cdot \stackrel{\to }{q}$$\vec{p} \cdot \vec{q}$

This proof requires one to recall the cosine rule of triangles. A simpler proof, that a student can easily derive, is given below.

easier proof

Bilinear Property : For any vector $\stackrel{\to }{p},\stackrel{\to }{q},\stackrel{\to }{r}\in V$$\vec{p} , \vec{q} , \vec{r} \in \mathbb{V}$ and $\lambda \in ℝ$$\lambda \in \mathbb{R}$
$\left(\lambda \stackrel{\to }{p}+\stackrel{\to }{q}\right)\cdot \stackrel{\to }{r}=\lambda \left(\stackrel{\to }{p}\cdot \stackrel{\to }{r}\right)+\left(\stackrel{\to }{q}\cdot \stackrel{\to }{r}\right)$$\left(\lambda \vec{p} + \vec{q}\right) \cdot \vec{r} = \lambda \left(\vec{p} \cdot \vec{r}\right) + \left(\vec{q} \cdot \vec{r}\right)$
This is explained and proven in properties of the dot product. For now, consider this to be true.

A vector $\stackrel{\to }{p}={p}_{x}i+{p}_{y}j+{p}_{z}k$$\vec{p} = {p}_{x} i + {p}_{y} j + {p}_{z} k$ is sum of scalar multiple of vectors. $i,j,k$$i , j , k$ are unit vectors, and the scalar multiples are ${p}_{x},{p}_{y},{p}_{z}$${p}_{x} , {p}_{y} , {p}_{z}$.

The same applies to $\stackrel{\to }{q}={q}_{x}i+{q}_{y}j+{q}_{z}k$$\vec{q} = {q}_{x} i + {q}_{y} j + {q}_{z} k$ Sum of multiple of vectors.

Proof for component form of vector dot product using bilinear property of dot product.

$\stackrel{\to }{p}\cdot \stackrel{\to }{q}$$\vec{p} \cdot \vec{q}$
$\quad \quad = \left({p}_{x} i + {p}_{y} j + {p}_{z} k\right) \cdot$
$\quad \quad \quad \quad \left({q}_{x} i + {q}_{y} j + {q}_{z} k\right)$
Apply bilinear property of dot product
$\quad \quad = {p}_{x} i \cdot \left({q}_{x} i + {q}_{y} j + {q}_{z} k\right)$
$\quad \quad \quad \quad + {p}_{y} j \cdot \left({q}_{x} i + {q}_{y} j + {q}_{z} k\right)$
$\quad \quad \quad \quad + {p}_{z} k \cdot \left({q}_{x} i + {q}_{y} j + {q}_{z} k\right)$
Apply $i\cdot i=i$$i \cdot i = i$,$j\cdot j=j$$j \cdot j = j$, $k\cdot k=k$$k \cdot k = k$
$i\cdot j=0$$i \cdot j = 0$, $j\cdot k=0$$j \cdot k = 0$, $k\cdot i=0$$k \cdot i = 0$

Given $\stackrel{\to }{p}=2i+1.2j-k$$\vec{p} = 2 i + 1.2 j - k$ and $\stackrel{\to }{q}=i-j+k$$\vec{q} = i - j + k$ what is $\stackrel{\to }{p}\cdot \stackrel{\to }{q}$$\vec{p} \cdot \vec{q}$?

The answer is '$-.2$$- .2$'

summary

Vector Dot Product in Component Form: For given two vectors $\stackrel{\to }{p}={p}_{x}i+{p}_{y}j+{p}_{z}k$$\vec{p} = {p}_{x} i + {p}_{y} j + {p}_{z} k$ and
$\stackrel{\to }{q}={q}_{x}i+{q}_{y}j+{q}_{z}k$$\vec{q} = {q}_{x} i + {q}_{y} j + {q}_{z} k$

$\stackrel{\to }{p}\cdot \stackrel{\to }{q}= {p}_{x}{q}_{x}+{p}_{y}{q}_{y}+{p}_{z}{q}_{z}$

For given two vectors in component forms, the dot product is the sum of product of corresponding components of the vectors.

Outline