Welcome to our community

Be a part of something great, join today!

[SOLVED] 311.1.5.14 Use vectors to describe this set as a line in R^4

karush

Well-known member
Jan 31, 2012
3,066
Screenshot 2020-12-23 at 11.41.43 AM.png
ok, just now looking at some examples of how to do this $x_4$ is just a row with all zeros
 

karush

Well-known member
Jan 31, 2012
3,066
ok don't see any takers on this one but here is a book example that might help, so we have....

$x_1+3x_4, \quad x_2=8+x_4, \quad x_3 =2-5x_4$ with $x_4$ free

$x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}
=\begin{bmatrix}3x_4\\8+x_4\\2-5x_4\\x_4\end{bmatrix}
=\begin{bmatrix}0\\8\\2\\0\end{bmatrix}= .....$

hopefully so far


Screenshot 2020-12-26 at 1.09.45 PM.png
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
Frankly it looks to me like you have no idea what you are supposed to be doing!

Yes, since we are told that "\(\displaystyle x_1= 3x_4\), \(\displaystyle x_2= 8+ x_4\), and \(\displaystyle x_3= 2- 5x_4\) we have immediately that \(\displaystyle \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}= \begin{bmatrix} 3x_4 \\ 8+ x_4 \\ 2- 5x_4 \\ x_4 \end{bmatrix}\).

But why then did you set \(\displaystyle x_4\) to 0?? The problem says that \(\displaystyle x_4\) is "free" which means that it can be any number. Saying that a number is "free" certainly does NOT mean that it is 0!

I would say that \(\displaystyle \begin{bmatrix} 3x_4 \\ 8+ x_4 \\ 2- 5x_4 \\ x_4 \end{bmatrix}\) is a perfectly good answer but some people might prefer to replace the "coordinate", \(\displaystyle x_4\) with the "parameter", t:
\(\displaystyle \begin{bmatrix} 3t \\ 8+ t \\ 2- 5t \\ t \end{bmatrix}\).

Some would prefer to write that as
\(\displaystyle \begin{bmatrix}0 \\ 8 \\ 2 \\ 0 \end{bmatrix}+\begin{bmatrix} 3 \\ 1 \\ -5 \\ 1 \end{bmatrix}t\).
 
Last edited:

karush

Well-known member
Jan 31, 2012
3,066
yes it is new material to me
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
Then you need to start by learning the basic definitions!