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Vector Dot Product

Vector Dot Product

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Vector Dot Product : Component Form


 »  sum of product of individual components
    →  `vec p cdot vec q = p_xq_x+p_yq_y+p_zq_z`
    →  `vec p = p_x i+p_yj+p_zk`
    →  `vec q = q_x i+q_yj+q_zk`

Vector Dot Product : Component Form

plain and simple summary

nub

plain and simple summary

nub

dummy

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

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trek

simple steps to build the foundation

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


Keep tapping on the content to continue learning.
Starting on learning "component form of vector dot product". ;; In this page, you will learn about the component form of the vector dot product.

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|`

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`

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

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`



           

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

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

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

The answer is '`-.2`'

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Given the two vectors p = p x i + p y j + p z k ;; vector q = q x i + q y j + q z k ;; what will be the vector dot product?
1
2
The answer is "magnitude of p multiplied magnitude of q multiplied cos theta". How will one compute the angle theta from the given component forms of the vectors p and q?
Consider this as triangle in coordinate plane. p = p x i + p y j + p z k ;; vector q = q x i + q y j + q z k ;; The triangle is made of 3 sides having scalar quantities p=magnitude of vector p, q = magnitude of vector q and r = vector q minus vector p. The cosine rule of triangle is applicable. r squared = p squared + q squared minus 2 p q cos theta. ;; With algebraic manipulations on this, we can derive that ;; cos theta = p x q x, + p y q y, + p z q z, divided by, magnitude of p magnitude of q. ;; Substituting the above in the vector dot product we get.;; vector p dot vector q = magnitude of p magnitude of q cos theta ;; equals p x q x, + p y q y, + p z q z.
That derives the component form of vector dot product as vector p dot vector q equals p x q x, + p y q y, + p z q 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 vectors p, q, r in vector space v, and lambda in real numbers ;; lambda times vector p + vector q, dot vector r ; equals ; lambda times vector p dot vector r + vector q dot vector r. ;; This is explained and proven in properties of the dot product. For now, consider this to be true.
A vector 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 vector 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.
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 vector p = p x i + p y j + p z k and vector q = q x i + q y j + q z k; vector p dot vector q = p x q x, + p y q y, + p z q z.
Given vector p and vector q, what is vector p dot vector q?
1
2
3
4
The answer is "minus point 2"

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