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Doubt regarding the projection along new basis

1

$begingroup$

I am trying to solve the following problem.
As per my understanding, the scalar projection of $vecp$ along $vecb1$ should be

$vecp$ . $vecb1$ / |$vecb1$|

Then why do we divide by |$vecb1$| twice? To get the correct answer we need to use the following formula instead, but why:

$vecp$ . $vecb1$ / |$vecb1$||$vecb1$|

linear-algebra vector-spaces vectors

share|cite|improve this question

asked Mar 30 at 17:07

TouchstoneTouchstone

1063

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct about the formula for the scalar projection, but the scalar projection is different from the coordinate.
  $endgroup$
  – jgon
  Mar 30 at 18:07
  

  
  
  
  

add a comment | 

1

$begingroup$

I am trying to solve the following problem.
As per my understanding, the scalar projection of $vecp$ along $vecb1$ should be

$vecp$ . $vecb1$ / |$vecb1$|

Then why do we divide by |$vecb1$| twice? To get the correct answer we need to use the following formula instead, but why:

$vecp$ . $vecb1$ / |$vecb1$||$vecb1$|

linear-algebra vector-spaces vectors

share|cite|improve this question

asked Mar 30 at 17:07

TouchstoneTouchstone

1063

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct about the formula for the scalar projection, but the scalar projection is different from the coordinate.
  $endgroup$
  – jgon
  Mar 30 at 18:07
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to solve the following problem.
As per my understanding, the scalar projection of $vecp$ along $vecb1$ should be

$vecp$ . $vecb1$ / |$vecb1$|

Then why do we divide by |$vecb1$| twice? To get the correct answer we need to use the following formula instead, but why:

$vecp$ . $vecb1$ / |$vecb1$||$vecb1$|

linear-algebra vector-spaces vectors

share|cite|improve this question

asked Mar 30 at 17:07

TouchstoneTouchstone

1063

$endgroup$

share|cite|improve this question

asked Mar 30 at 17:07

TouchstoneTouchstone

1063

share|cite|improve this question

asked Mar 30 at 17:07

TouchstoneTouchstone

1063

asked Mar 30 at 17:07

TouchstoneTouchstone

1063

asked Mar 30 at 17:07

TouchstoneTouchstone

1063

1 Answer
1

active

oldest

votes

0

$begingroup$

Rather than using the formula above, I would say:

$v = c_1 b_1 + c_2 b_2\
c_1 + c_2 = 5\
c_1 - c_2 = -1$

And now I have a rather simple system of equations.

At a deeper level of understanding, I might say:

$T = beginbmatrix 1 & 1 \ 1 & -1endbmatrix$ would be a transformation from the basis $B$ to the standard basis.

In which chase $T^-1v$ would transform v in the standard basis to the basis $B.$

As for what you have in your formula above.

We take $b_1$ and we normalize it.

i.e. $u_1 = frac b_1$

Then we calculate the dot product.

$vcdot u_1$ giving us a magnitude in this direction. Multiply by the direction vector

$(vcdot u_1) u_1$

Substituting back

$(vcdot u_1) u_1$ = $(vcdot frac b_1) frac b_1 = frac vcdot b_1b_1$

share|cite|improve this answer

answered Mar 30 at 17:22

Doug MDoug M

1,848412

$endgroup$

add a comment | 

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0

$begingroup$

Rather than using the formula above, I would say:

$v = c_1 b_1 + c_2 b_2\
c_1 + c_2 = 5\
c_1 - c_2 = -1$

And now I have a rather simple system of equations.

At a deeper level of understanding, I might say:

$T = beginbmatrix 1 & 1 \ 1 & -1endbmatrix$ would be a transformation from the basis $B$ to the standard basis.

In which chase $T^-1v$ would transform v in the standard basis to the basis $B.$

As for what you have in your formula above.

We take $b_1$ and we normalize it.

i.e. $u_1 = frac b_1$

Then we calculate the dot product.

$vcdot u_1$ giving us a magnitude in this direction. Multiply by the direction vector

$(vcdot u_1) u_1$

Substituting back

$(vcdot u_1) u_1$ = $(vcdot frac b_1) frac b_1 = frac vcdot b_1b_1$

share|cite|improve this answer

answered Mar 30 at 17:22

Doug MDoug M

1,848412

$endgroup$

add a comment | 

0

$begingroup$

Rather than using the formula above, I would say:

$v = c_1 b_1 + c_2 b_2\
c_1 + c_2 = 5\
c_1 - c_2 = -1$

And now I have a rather simple system of equations.

At a deeper level of understanding, I might say:

$T = beginbmatrix 1 & 1 \ 1 & -1endbmatrix$ would be a transformation from the basis $B$ to the standard basis.

In which chase $T^-1v$ would transform v in the standard basis to the basis $B.$

As for what you have in your formula above.

We take $b_1$ and we normalize it.

i.e. $u_1 = frac b_1$

Then we calculate the dot product.

$vcdot u_1$ giving us a magnitude in this direction. Multiply by the direction vector

$(vcdot u_1) u_1$

Substituting back

$(vcdot u_1) u_1$ = $(vcdot frac b_1) frac b_1 = frac vcdot b_1b_1$

share|cite|improve this answer

answered Mar 30 at 17:22

Doug MDoug M

1,848412

$endgroup$

add a comment | 

$begingroup$

Rather than using the formula above, I would say:

$v = c_1 b_1 + c_2 b_2\
c_1 + c_2 = 5\
c_1 - c_2 = -1$

And now I have a rather simple system of equations.

At a deeper level of understanding, I might say:

$T = beginbmatrix 1 & 1 \ 1 & -1endbmatrix$ would be a transformation from the basis $B$ to the standard basis.

In which chase $T^-1v$ would transform v in the standard basis to the basis $B.$

As for what you have in your formula above.

We take $b_1$ and we normalize it.

i.e. $u_1 = frac b_1$

Then we calculate the dot product.

$vcdot u_1$ giving us a magnitude in this direction. Multiply by the direction vector

$(vcdot u_1) u_1$

Substituting back

$(vcdot u_1) u_1$ = $(vcdot frac b_1) frac b_1 = frac vcdot b_1b_1$

share|cite|improve this answer

answered Mar 30 at 17:22

Doug MDoug M

1,848412

$endgroup$

share|cite|improve this answer

answered Mar 30 at 17:22

Doug MDoug M

1,848412

answered Mar 30 at 17:22

Doug MDoug M

1,848412

answered Mar 30 at 17:22

Doug MDoug M

1,848412