Vector Algebra : Vector Dot Product : Component Form

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Vector Dot Product : Component Form

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Overview

Vector Dot Product : Component Form

» sum of product of individual components
→ $\vec{p} \cdot \vec{q} = p_{x} q_{x} + p_{y} q_{y} + p_{z} q_{z}$ $vec p cdot vec q = p_xq_x+p_yq_y+p_zq_z$

simple in component form

Given the two vectors
$\vec{p} = p_{x} i + p_{y} j + p_{z} k$ $vec p = p_x i+p_yj+p_zk$

$\vec{q} = q_{x} i + q_{y} j + q_{z} k$ $vec q = q_x i+q_yj+q_zk$ The vector dot product is defined as $| p | | q | \cos θ$ $|p||q|cos theta$

How will one compute the angle $θ$ $theta$ from the given component forms of vectors?

Consider this as triangle in coordinate plane. The triangle is made of 3 sides having scalar quantities $p = | \vec{p} |$ $p=|vec p|$ , $q = | \vec{q} |$ $q=|vec q|$ and $r = | \vec{q} - \vec{p} |$ $r = |vec q-vec p|$ . The cosine rule of triangle is applicable.
$r^{2} = p^{2} + q^{2} - 2 p q \cos θ$ $r^2=p^2+q^2-2pq cos theta$

$r^{2} = {(q_{x} - p_{x})}^{2} + {(q_{y} - p_{y})}^{2} + {(q_{z} - p_{z})}^{2}$ $r^2 = (q_x-p_x)^2 + (q_y-p_y)^2 + (q_z-p_z)^2$ ,
$p^{2} = p_{x}^{2} + p_{y}^{2} + p_{z}^{2}$ $p^2 = p_x^2 + p_y^2 + p_z^2$ ,
$q^{2} = q_{x}^{2} + q_{y}^{2} + q_{z}^{2}$ $q^2 = q_x^2 + q_y^2 + q_z^2$ ,

With algebraic manipulations on this, we can derive that
$\cos θ = \frac{p_{x} q_{x} + p_{y} q_{y} + p_{z} q_{z}}{| p | | q |}$ $cos theta = (p_xq_x+p_yq_y+p_zq_z)/(|p||q|)$
Substituting the above in the vector dot product we get.
$\vec{p} \cdot \vec{q}$ $vec p cdot vec q$
$= | p | | q | \cos θ$ $quad quad = |p||q|cos theta$
$= | p | | q | \frac{p_{x} q_{x} + p_{y} q_{y} + p_{z} q_{z}}{| p | | q |}$ $quad quad = |p||q|(p_xq_x+p_yq_y+p_zq_z)/(|p||q|)$
$= p_{x} q_{x} + p_{y} q_{y} + p_{z} q_{z}$ $quad quad = p_xq_x+p_yq_y+p_zq_z$

That derives the component form of vector dot product as
$\vec{p} \cdot \vec{q}$ $vec p cdot vec q$
$= p_{x} q_{x} + p_{y} q_{y} + p_{z} q_{z}$ $quad quad = p_xq_x+p_yq_y+p_zq_z$

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 $\vec{p}, \vec{q}, \vec{r} \in V$ $vec p, vec q, vec r in bbb V$ and $λ \in ℝ$ $lambda in RR$
$(λ \vec{p} + \vec{q}) \cdot \vec{r} = λ (\vec{p} \cdot \vec{r}) + (\vec{q} \cdot \vec{r})$ $(lambda vec p + vec q) cdot vec r = lambda (vec p cdot vec r) + (vec q cdot vec r)$
This is explained and proven in properties of the dot product. For now, consider this to be true.

A vector $\vec{p} = p_{x} i + p_{y} j + p_{z} k$ $vec p = p_x i + p_y j + p_z k$ 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 $\vec{q} = q_{x} i + q_{y} j + q_{z} k$ $vec q = q_x i + q_y j + q_z k$ Sum of multiple of vectors.

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

$\vec{p} \cdot \vec{q}$ $vec p cdot vec q$
$= (p_{x} i + p_{y} j + p_{z} k) \cdot$ $quad quad = (p_x i + p_y j + p_z k) cdot$
$(q_{x} i + q_{y} j + q_{z} k)$ $quad quad quad quad (q_x i + q_y j + q_z k)$
Apply bilinear property of dot product
$= p_{x} i \cdot (q_{x} i + q_{y} j + q_{z} k)$ $quad quad = p_x i cdot (q_x i + q_y j + q_z k)$
$+ p_{y} j \cdot (q_{x} i + q_{y} j + q_{z} k)$ $quad quad quad quad + p_y j cdot (q_x i + q_y j + q_z k)$
$+ p_{z} k \cdot (q_{x} i + q_{y} j + q_{z} k)$ $quad quad quad quad + p_z k cdot (q_x i + q_y j + q_z k)$
Apply $i \cdot i = i$ $i cdot i = i$ , $j \cdot j = j$ $j cdot j = j$ , $k \cdot k = k$ $k cdot k = k$
$i \cdot j = 0$ $i cdot j = 0$ , $j \cdot k = 0$ $j cdot k = 0$ , $k \cdot i = 0$ $k cdot i = 0$
$= p_{x} q_{x} + p_{y} q_{y} + p_{z} q_{z}$ $quad quad = p_xq_x+p_yq_y+p_zq_z$

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

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

summary

Vector Dot Product in Component Form: For given two vectors $\vec{p} = p_{x} i + p_{y} j + p_{z} k$ $vec p = p_x i+p_yj+p_zk$ and
$\vec{q} = q_{x} i + q_{y} j + q_{z} k$ $vec q = q_x i+q_yj+q_zk$

$\vec{p} \cdot \vec{q} = p_{x} q_{x} + p_{y} q_{y} + p_{z} q_{z}$ $vec p cdot vec q = p_xq_x+p_yq_y+p_zq_z$

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

Outline

The outline of material to learn vector-algebra is as follows.

Note: Click here for detailed outline of vector-algebra.

• Introduction to Vectors

→ Introducing Vectors

→ Representation of Vectors

• Basic Properties of Vectors

→ Magnitude of Vectors

→ Types of Vectors

→ Properties of Magnitude

• Vectors & Coordinate Geometry

→ Vectors & Coordinate Geometry

→ Position Vector of a point

→ Directional Cosine

• Role of Direction in Vector Arithmetics

→ Vector Arithmetics

→ Understanding Direction of Vectors

• Vector Addition

→ Vector Additin : First Principles

→ Vector Addition : Component Form

→ Triangular Law

→ Parallelogram Law

• Multiplication of Vector by Scalar

→ Scalar Multiplication

→ Standard Unit Vectors

→ Vector as Sum of Vectors

→ Vector Component Form

• Vector Dot Product

→ Introduction to Vector Multiplication

→ Cause-Effect-Relation

→ Dot Product : First Principles

→ Dot Product : Projection Form

→ Dot Product : Component Form

→ Dot Product With Direction

• Vector Cross Product

→ Vector Multiplication : Cross Product

→ Cross Product : First Principles

→ Cross Product : Area of Parallelogram

→ Cross Product : Component Form

→ Cross Product : Direction Removed