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Equation with norm

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote an unsolved question from MHF posted by user Civy71 on February 18th, 2013
Solve the equation 5x - ||v|| v = ||w||(w-5x) for x with v = (1, 2, -4, 0) and w = (-3, 5, 1, 1)
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
We have:

$
\begin{aligned}5x-||v||v=||w||(w-5x)&\Leftrightarrow 5x+5||w||x=||w||w+||v||v\\&\Leftrightarrow 5(1+||w||)x=||w||w+||v||v\\&\Leftrightarrow x=\dfrac{1}{5(1+||w||)}(||w||w+||v||v)
\end{aligned}
$

Now, substitute $v=(1, 2, -4, 0)$, $w=(-3, 5, 1, 1 )$, $||w||=6$, $||v||=\sqrt{21}$.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Personally, I would have immediately substituted the given values for v and w.

5x - ||v|| v = ||w||(w-5x) for x with v = (1, 2, -4, 0) and w = (-3, 5, 1, 1)
so [tex]||v||= \sqrt{1+ 4+ 16}= \sqrt{21}[/tex] and [tex]||w||= \sqrt{9+ 25+ 1+ 1}= \sqrt{38}[/tex]

Let x= (w, x, y, z) so the equation becomes
[tex](5w, 5x, 5y, 5z)- (\sqrt{21}, 2\sqrt{21}, -4\sqrt{21}, 0)= (-3\sqrt{38}, 5\sqrt{38}, \sqrt{38}, \sqrt{38})- (5\sqrt{38}w, 5\sqrt{38}x, 5\sqrt{38}y, 5\sqrt{38}z)[/tex]
which gives the four numeric equations
[tex]5w- \sqrt{21}= -3\sqrt{38}- 5\sqrt{38}w[/tex]
[tex]5x- 2\sqrt{21}= 5\sqrt{38}- 5\sqrt{38}x[/tex]
[tex]5y+ 4\sqrt{21}= \sqrt{38}- 5\sqrt{38}y[/tex]
[tex]5z= \sqrt{38}- 5\sqrt{38}z[/tex]
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Personally, I would have immediately substituted the given values for v and w.
All right, no problem with that. My preference was for the sake of generalization.