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Projections: Finding Components and Projections of Vectors, Lecture notes of Physics

University of Pittsburgh (Pitt) - Medical Center-Health System Physics

How to find the projection and component of a vector in the direction of another vector. It includes formulas, examples, and the concept of orthogonal projections.

Typology: Lecture notes

2021/2022

Uploaded on 09/27/2022

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7.7 Projections P. Danziger

Components and Projections





AAAAAA





Avu





projvu

Given two vectors uand v, we can ask how far we

will go in the direction of vwhen we travel along

The distance we travel in the direction of v, while

traversing uis called the component of uwith

respect to vand is denoted compvu.

The vector parallel to v, with magnitude compvu,

in the direction of vis called the projection of u

onto vand is denoted projvu.

So, compvu=||projvu||

Note projvuis a vector and compvuis a scalar.

From the picture compvu=||u||cos θ

Discover Lecture notes of Physics University of Pittsburgh (Pitt) - Medical Center-Health System

Partial preview of the text

Download Projections: Finding Components and Projections of Vectors and more Lecture notes Physics in PDF only on Docsity!

Components and Projections

AA AA AA

u^ A v

projvu

Given two vectors u and v, we can ask how far we

will go in the direction of v when we travel along

u.

The distance we travel in the direction of v, while

traversing u is called the component of u with

respect to v and is denoted compvu.

The vector parallel to v, with magnitude compvu,

in the direction of v is called the projection of u

onto v and is denoted projvu.

So, compvu = ||projvu||

Note projvu is a vector and compvu is a scalar.

From the picture compvu = ||u|| cos θ

We wish to find a formula for the projection of u

onto v.

Consider u · v = ||u||||v|| cos θ

Thus ||u|| cos θ = u ||^ v·^ ||v

So compvu = u ||^ v·^ ||v

The unit vector in the same direction as v is given

by ||vv||. So

projvu =

(

u · v

||v||^2

)

v

4. Find the components of u = i + 3j − 2 k in the

directions i, j and k.

u · i = 1, u · j = 3, u · k = − 2 ,

||i|| = ||j|| = ||k|| = 1

compiu = 1, compju = 3, compku = − 2.

So the use of the term component is justified in this context.

Indeed, coordinate axes are arbitrarily chosen and are subject to change.

If u is a new coordinate vector given in terms of

the old set then compuw gives the component

of the vector w in the new coordinate system.

Example 2

If coordinates in the plane are rotated by 45o,

the vector i is mapped to u = √^12 i + √^12 j, and

the vector j is mapped to v = − √^12 i+ √^12 j. Find

the components of w = 2i − 5 j with respect

to the new coordinate vectors u and v. i.e. Express w in terms of u and v.

w @@ w

i

j 6 ^6 -- v@@I @@u

w · u =

√−^3

, w · v =

. ||u|| = ||v|| = 1

compuw = √−^3

, compvw = √−^7

and

w =

u +

v

Example 3

Express u = 2i+4j+2k as a sum of vectors parallel

and perpendicular to v = i + 2j − k.

u·v = 2+8−2 = 8, ||v||^2 =

(√ 12 + 2^2 + 1^2

) 2 = 6

u|| = projvu =

(

u · v

||v||^2

)

v =

(i + 2j − k)

u⊥ = u − projvu

= (2i + 4j + 2k) − 43 (i + 2j − k)

( 2 − (^43)

)

i +

( 4 − (^83)

)

j +

( 2 + (^43)

)

k

= 6 − 3 4 i + 123 − 8 j + 6+4 3 k

= 23 i + 43 j + 103 k

= 23 (i + 2j + 5k)

Check

u|| · u⊥ =

( 2

3 (i^ + 2j^ + 5k)

) ·

( 4

3 (i^ + 2j^ −^ k)

)

Projections: Finding Components and Projections of Vectors, Lecture notes of Physics

Related documents

Partial preview of the text

Download Projections: Finding Components and Projections of Vectors and more Lecture notes Physics in PDF only on Docsity!

Components and Projections

u^ A v

projvu

Given two vectors u and v, we can ask how far we

will go in the direction of v when we travel along

u.

The distance we travel in the direction of v, while

traversing u is called the component of u with

respect to v and is denoted compvu.

The vector parallel to v, with magnitude compvu,

in the direction of v is called the projection of u

onto v and is denoted projvu.

So, compvu = ||projvu||

Note projvu is a vector and compvu is a scalar.

From the picture compvu = ||u|| cos θ

We wish to find a formula for the projection of u

onto v.

Consider u · v = ||u||||v|| cos θ

Thus ||u|| cos θ = u ||^ v·^ ||v

So compvu = u ||^ v·^ ||v

The unit vector in the same direction as v is given

by ||vv||. So

projvu =

u · v

||v||^2

v

4. Find the components of u = i + 3j − 2 k in the

directions i, j and k.

u · i = 1, u · j = 3, u · k = − 2 ,

||i|| = ||j|| = ||k|| = 1

compiu = 1, compju = 3, compku = − 2.

the old set then compuw gives the component

the vector i is mapped to u = √^12 i + √^12 j, and

the vector j is mapped to v = − √^12 i+ √^12 j. Find

the components of w = 2i − 5 j with respect

w @@ w

i

j 6 ^6 -- v@@I @@u

w · u =

√−^3

, w · v =

. ||u|| = ||v|| = 1

compuw = √−^3

, compvw = √−^7

w =

u +

v

Express u = 2i+4j+2k as a sum of vectors parallel

and perpendicular to v = i + 2j − k.

u·v = 2+8−2 = 8, ||v||^2 =

u|| = projvu =

u · v

||v||^2

v =

(i + 2j − k)

u⊥ = u − projvu

= (2i + 4j + 2k) − 43 (i + 2j − k)

i +

j +

k

= 6 − 3 4 i + 123 − 8 j + 6+4 3 k

= 23 i + 43 j + 103 k

= 23 (i + 2j + 5k)

u|| · u⊥ =

3 (i^ + 2j^ + 5k)

3 (i^ + 2j^ −^ k)

= 89 ((i + 2j + 5k) · (i + 2j − k))

So u|| and u⊥ are orthogonal.

u^ A v

j 6 ^6 -- v@@I @@u