MHB *412 what value(s) of h is b in plane spanned

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The value of \( h \) for which vector \( b \) lies in the plane spanned by vectors \( a_1 \) and \( a_2 \) is definitively \( h = 3 \). The vectors are defined as \( a_1 = \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} \), \( a_2 = \begin{bmatrix} -5 \\ -8 \\ 2 \end{bmatrix} \), and \( b = \begin{bmatrix} 3 \\ -5 \\ h \end{bmatrix} \). The solution involves setting up the augmented matrix and reducing it to row echelon form, leading to the conclusion that \( v = -7 \) and \( w = -2 \) are the coefficients for the linear combination.

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karush
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For what value(s) of $h$ is b in the plane spanned by $a_1$ and $a_2$
$$a_1=\left[\begin{array}{r} 1\\3\\ -1 \end{array}\right],
a_2=\left[\begin{array}{r} -5\\-8\\2 \end{array}\right],
b =\left[\begin{array}{r} 3\\-5\\ \color{red}{h} \end{array}\right]$$

ok this should be obvious but I don't see it..
 
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karush said:
For what value(s) of $h$ is b in the plane spanned by $a_1$ and $a_2$
$$a_1=\left[\begin{array}{r} 1\\3\\ -1 \end{array}\right],
a_2=\left[\begin{array}{r} -5\\-8\\2 \end{array}\right],
b =\left[\begin{array}{r} 3\\-5\\ \color{red}{h} \end{array}\right]$$

ok this should be obvious but I don't see it..
Hint: If b is in the plane formed by a_1 and a_2 then it has to be a linear combination of a_1 and a_2. ie. [math]b = v a_1 + w a_2[/math] for some v, w constants.

Can you finish?

-Dan
 
topsquark said:
Hint: If b is in the plane formed by a_1 and a_2 then it has to be a linear combination of a_1 and a_2. ie. [math]b = v a_1 + w a_2[/math] for some v, w constants.

Can you finish?

-Dan

$\left[\begin{array}{r} 1\\3\\ -1 \end{array}\right]v+
\left[\begin{array}{r} -5\\-8\\2 \end{array}\right]w
=\left[\begin{array}{r} 3\\-5\\ \color{red}{h} \end{array}\right]$
so then the augmented matrix would be
$\left[\begin{array}{rr|r}1 & -5 & 3 \\ 3 & -8 & -5 \\ -1 & 2 & h \end{array}\right]$
then RREF
$\left[ \begin{array}{cc|c} 1 & 0 & -7 \\0 & 1 & -2 \\ 0 & 0 & h - 3 \end{array} \right]$
so $h=3$ following would be $v=7$ and $w=2$

hopefully...
 
Last edited:
karush said:
$\left[\begin{array}{r} 1\\3\\ -1 \end{array}\right]v+
\left[\begin{array}{r} -5\\-8\\2 \end{array}\right]w
=\left[\begin{array}{r} 3\\-5\\ \color{red}{h} \end{array}\right]$
so then the augmented matrix would be
$\left[\begin{array}{rr|r}1 & -5 & 3 \\ 3 & -8 & -5 \\ -1 & 2 & h \end{array}\right]$
then RREF
$\left[ \begin{array}{cc|c} 1 & 0 & -7 \\0 & 1 & -2 \\ 0 & 0 & h - 3 \end{array} \right]$
so $h=3$ following would be $v=7$ and $w=2$

hopefully...
I didn't go looking for it but somehow you are off by a sign. v = -7 and w = -2 and h = 3 is the solution.

-Dan
 
ok I see

however the OP only asked for h

mahalo
 
karush said:
ok I see

however the OP only asked for h

mahalo
I know. It was just an FYI.

-Dan
 
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