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How can I visualize $cmathbfv + (1 - c)mathbfw$?

Geometric/Visual Solution - Shortest Vector for which Dot Product = x + 2y = 5. (Strang P21 1.2.26)Row and Column Picture of a 3 x 3 Singular Matrix (Strang P43, 2.1.32)Column and Row Picture for Singular System of 100 Equations (Strang P55, 2.2.32)Columns of A - Linear independence, span, rank. [GStrang P180, 3.5.19]Do the vectors not in the column space form a subspace? [GStrang P130 3.1.27]Problem with alternate solution — Equation of plane through point and containing intersection line of two planes [Stewart P $803, 12.5.37$]How is the “two intersecting lines” picture of a system of two linear equations related to the “sum of two vectors” picture?How Can the Maximum Number of Moves Needed to Solve a 3x3x3 Rubik's Cube Be Simply ProvenFind the line through the origin farthest to the points (1,1), (1,−1), and (−1,−1).Sampling points on a plane defined by three points

5

2

$begingroup$

Draw the line of all combinations that has $cmathbfv + dmathbfw$ and $c + d = 1$.

Solution: All combinations with $c + d = 1$ are on the line that passes through $mathbfv$ and $mathbfw$.

1. $cmathbfv + dmathbfw quad & quad c + d = 1 implies cmathbfv + (1 - c)mathbfw$.
   
   But how do I draw $(1 - c)mathbfw$?




2. My $c = 1.5$ sketch fits the solution. But why doesn't my sketch for $c = -0.5$?

I additionally tried $cmathbfv + (1 - c)mathbfw = color#318CE7c(mathbfv - mathbfw) + mathbfw $ in picture 2 but it doesn't agree with solution?

3. How can I get the solution algebraically, without pictures?

linear-algebra vector-spaces

share|cite|improve this question

edited Mar 15 at 3:43

Adopt. Don't reproduce.

asked Jul 31 '13 at 7:52

Adopt. Don't reproduce.Adopt. Don't reproduce.

264

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think in picture 2 if you shifted the initial point of c(v-w) to origin then your observation will agree with the result..
  $endgroup$
  – Ritu
  Feb 20 '15 at 7:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And for $c=-0.5$, start $-0.5v$ from origin...
  $endgroup$
  – Ritu
  Feb 20 '15 at 7:16
  

  
  
  
  

add a comment | 

5

2

$begingroup$

Draw the line of all combinations that has $cmathbfv + dmathbfw$ and $c + d = 1$.

Solution: All combinations with $c + d = 1$ are on the line that passes through $mathbfv$ and $mathbfw$.

1. $cmathbfv + dmathbfw quad & quad c + d = 1 implies cmathbfv + (1 - c)mathbfw$.
   
   But how do I draw $(1 - c)mathbfw$?




2. My $c = 1.5$ sketch fits the solution. But why doesn't my sketch for $c = -0.5$?

I additionally tried $cmathbfv + (1 - c)mathbfw = color#318CE7c(mathbfv - mathbfw) + mathbfw $ in picture 2 but it doesn't agree with solution?

3. How can I get the solution algebraically, without pictures?

linear-algebra vector-spaces

share|cite|improve this question

edited Mar 15 at 3:43

Adopt. Don't reproduce.

asked Jul 31 '13 at 7:52

Adopt. Don't reproduce.Adopt. Don't reproduce.

264

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think in picture 2 if you shifted the initial point of c(v-w) to origin then your observation will agree with the result..
  $endgroup$
  – Ritu
  Feb 20 '15 at 7:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And for $c=-0.5$, start $-0.5v$ from origin...
  $endgroup$
  – Ritu
  Feb 20 '15 at 7:16
  

  
  
  
  

add a comment | 

$begingroup$

Draw the line of all combinations that has $cmathbfv + dmathbfw$ and $c + d = 1$.

Solution: All combinations with $c + d = 1$ are on the line that passes through $mathbfv$ and $mathbfw$.

1. $cmathbfv + dmathbfw quad & quad c + d = 1 implies cmathbfv + (1 - c)mathbfw$.
   
   But how do I draw $(1 - c)mathbfw$?




2. My $c = 1.5$ sketch fits the solution. But why doesn't my sketch for $c = -0.5$?

I additionally tried $cmathbfv + (1 - c)mathbfw = color#318CE7c(mathbfv - mathbfw) + mathbfw $ in picture 2 but it doesn't agree with solution?

3. How can I get the solution algebraically, without pictures?

linear-algebra vector-spaces

share|cite|improve this question

edited Mar 15 at 3:43

Adopt. Don't reproduce.

asked Jul 31 '13 at 7:52

Adopt. Don't reproduce.Adopt. Don't reproduce.

264

$endgroup$

share|cite|improve this question

edited Mar 15 at 3:43

Adopt. Don't reproduce.

asked Jul 31 '13 at 7:52

Adopt. Don't reproduce.Adopt. Don't reproduce.

264

share|cite|improve this question

edited Mar 15 at 3:43

Adopt. Don't reproduce.

asked Jul 31 '13 at 7:52

Adopt. Don't reproduce.Adopt. Don't reproduce.

264

asked Jul 31 '13 at 7:52

Adopt. Don't reproduce.Adopt. Don't reproduce.

264

asked Jul 31 '13 at 7:52

Adopt. Don't reproduce.Adopt. Don't reproduce.

264

2 Answers
2

active

oldest

votes

2

$begingroup$

Maybe this little illustration will help too (sorry for the quality). The problem in issue#2 is the wrong addition of vectors.

share|cite|improve this answer

answered Jul 31 '13 at 8:42

EvgenyEvgeny

4,71021022

$endgroup$

add a comment | 

1

$begingroup$

Suppose $AB$ and $AC$ are your vectors $v$ and $w$, respectively. Then $AD$ is your convex combination $uv+(1-u)w$, where $uv$ is $AF$, and $AE$ is $(1-u)w$. The similarities between the triangles $BFD$, $EDC$ and $ABC$ explains it.

The fact that $BD+DC=BC$ is related to $(1-u)+u=1$, because you get $BD/BC+DC/BC=1$. Now, the triangles similarities gives you that $BD/BC=FD/AC=AE/AC=1-u$ and $DC/BC=ED/AB=AF/AB=u$.

share|cite|improve this answer

edited Jul 31 '13 at 8:42

answered Jul 31 '13 at 8:27

OR.OR.

5,19021437

$endgroup$

add a comment | 

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2

$begingroup$

Maybe this little illustration will help too (sorry for the quality). The problem in issue#2 is the wrong addition of vectors.

share|cite|improve this answer

answered Jul 31 '13 at 8:42

EvgenyEvgeny

4,71021022

$endgroup$

add a comment | 

2

$begingroup$

Maybe this little illustration will help too (sorry for the quality). The problem in issue#2 is the wrong addition of vectors.

share|cite|improve this answer

answered Jul 31 '13 at 8:42

EvgenyEvgeny

4,71021022

$endgroup$

add a comment | 

$begingroup$

Maybe this little illustration will help too (sorry for the quality). The problem in issue#2 is the wrong addition of vectors.

share|cite|improve this answer

answered Jul 31 '13 at 8:42

EvgenyEvgeny

4,71021022

$endgroup$

share|cite|improve this answer

answered Jul 31 '13 at 8:42

EvgenyEvgeny

4,71021022

answered Jul 31 '13 at 8:42

EvgenyEvgeny

4,71021022

answered Jul 31 '13 at 8:42

EvgenyEvgeny

4,71021022

1

$begingroup$

Suppose $AB$ and $AC$ are your vectors $v$ and $w$, respectively. Then $AD$ is your convex combination $uv+(1-u)w$, where $uv$ is $AF$, and $AE$ is $(1-u)w$. The similarities between the triangles $BFD$, $EDC$ and $ABC$ explains it.

The fact that $BD+DC=BC$ is related to $(1-u)+u=1$, because you get $BD/BC+DC/BC=1$. Now, the triangles similarities gives you that $BD/BC=FD/AC=AE/AC=1-u$ and $DC/BC=ED/AB=AF/AB=u$.

share|cite|improve this answer

edited Jul 31 '13 at 8:42

answered Jul 31 '13 at 8:27

OR.OR.

5,19021437

$endgroup$

add a comment | 

1

$begingroup$

Suppose $AB$ and $AC$ are your vectors $v$ and $w$, respectively. Then $AD$ is your convex combination $uv+(1-u)w$, where $uv$ is $AF$, and $AE$ is $(1-u)w$. The similarities between the triangles $BFD$, $EDC$ and $ABC$ explains it.

The fact that $BD+DC=BC$ is related to $(1-u)+u=1$, because you get $BD/BC+DC/BC=1$. Now, the triangles similarities gives you that $BD/BC=FD/AC=AE/AC=1-u$ and $DC/BC=ED/AB=AF/AB=u$.

share|cite|improve this answer

edited Jul 31 '13 at 8:42

answered Jul 31 '13 at 8:27

OR.OR.

5,19021437

$endgroup$

add a comment | 

$begingroup$

Suppose $AB$ and $AC$ are your vectors $v$ and $w$, respectively. Then $AD$ is your convex combination $uv+(1-u)w$, where $uv$ is $AF$, and $AE$ is $(1-u)w$. The similarities between the triangles $BFD$, $EDC$ and $ABC$ explains it.

The fact that $BD+DC=BC$ is related to $(1-u)+u=1$, because you get $BD/BC+DC/BC=1$. Now, the triangles similarities gives you that $BD/BC=FD/AC=AE/AC=1-u$ and $DC/BC=ED/AB=AF/AB=u$.

share|cite|improve this answer

edited Jul 31 '13 at 8:42

answered Jul 31 '13 at 8:27

OR.OR.

5,19021437

$endgroup$

share|cite|improve this answer

edited Jul 31 '13 at 8:42

answered Jul 31 '13 at 8:27

OR.OR.

5,19021437

answered Jul 31 '13 at 8:27

OR.OR.

5,19021437

answered Jul 31 '13 at 8:27

OR.OR.

5,19021437