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### 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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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. 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
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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?
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The answer is "minus point 2"

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