Server Error

Server Not Reachable.

This may be due to your internet connection or the nubtrek server is offline.

Thought-Process to Discover Knowledge

Welcome to nubtrek.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue
mathsVector AlgebraVector Dot Product

Vector Dot Product : Component Form

In this page, you will learn about the component form of the vector dot product.

click on the content to continue..

Given the two vectors
`vec p = p_x i+p_yj+p_zk`

`vec q = q_x i+q_yj+q_zk`
what will be the vector dot product?

  • `|p||q|cos theta`
  • `|p||q|cos theta`
  • `|p||q|`

answer is `|p||q|cos theta`.

How will one compute the angle `theta` from the given component forms of vectors
`vec p = p_x i+p_yj+p_zk`
`vec q = q_x i+q_yj+q_zk`

Consider this as triangle in coordinate plane.
`vec p = p_x i+p_yj+p_zk`
`vec q = q_x i+q_yj+q_zk`.vector dot product formula derivation The triangle is made of 3 sides having scalar quantities `p=|vec p|`, `q=|vec q|` and `r = |vec q-vec p|`. The cosine rule of triangle is applicable.
`r^2=p^2+q^2-2pq cos theta`

With algebraic manipulations on this, we can derive that
`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`
 `quad quad = |p||q|cos theta`
 `quad quad = |p||q|(p_xq_x+p_yq_y+p_zq_z)/(|p||q|)`
 `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`
 `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 in the coming pages.

Bilinear Property : For any vector `vec p, vec q, vec r in bbb V` and `lambda in RR`
`(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` is sum of scalar multiple of vectors. `i, j, k` are unit vectors, and the scalar multiples are `p_x, p_y, p_z`.

The same applies to `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`
`quad quad = (p_x i + p_y j + p_z k) cdot`
`quad quad quad quad (q_x i + q_y j + q_z k)`
Apply bilinear property of dot product
`quad quad = p_x i 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)`
`quad quad quad quad + p_z k cdot (q_x i + q_y j + q_z k)`
Apply `i cdot i = i`,`j cdot j = j`, `k cdot k = k`
`i cdot j = 0`, `j cdot k = 0`, `k cdot i = 0`
`quad quad = 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.

Vector Dot Product in Component Form: For given two vectors `vec p = p_x i+p_yj+p_zk` and
`vec q = q_x i+q_yj+q_zk`

`vec p cdot vec q = p_xq_x+p_yq_y+p_zq_z`

Solved Exercise Problem:

Given `vec p = 2i+1.2j-k` and `vec q = i-j+k` what is `vec p cdot vec q`?

  • `2.2`
  • `.2`
  • `-.2`
  • `-.2`
  • `4.2`

The answer is '`-.2`'

slide-show version coming soon