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 April 6th, 2015, 02:31 AM #1 Senior Member   Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4 Finding a vector that is parallel to a plane Hello, If I am given a vector u=8i+6j+7k. and I need to find another vector w that is parallel to the XZ plane and also perpendicular to the given vector u. I know that the XZ plane means that y=0. I also understand that for the vector to be perpendicular I need to find that the dot product should be 0. But I can't find out what to do exactly to find that vector. Thanks for your help
April 6th, 2015, 03:57 AM   #2
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Joined: Dec 2006
From: Lexington, MA

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Hello, noobinmath!

Quote:
 Given vector $\vec{u}\,=\,8i\,+\,6j\,+\,7k.$ Find $\vec{w}$ that is parallel to the $xz$-plane and also perpendicular to the given vector $\vec{u}.$

Your reasoning is correct.

$\text{Let }\vec w \:=\:\langle a,\,b,\,c\rangle$

$\text{Since }\vec w\text{ is parallel to the }xz\text{-plane, }b\,=\,0.$
$\text{W\!e have: }\:\vec w \:=\:\langle a,\,0,\,c\rangle$

$\text{Since }\vec u\,\perp\,\vec w,\;\vec u\,\cdot\,\vec w \,=\,0.$
$\text{W\!e have: }\:\langle 8,\,6,\,7\rangle\,\cdot\,\langle a,\,0,\,c\rangle \:=\:0$
$\;\;\;8a\,+\,0\,+\,7c\:=\:0 \;\;\;\Rightarrow\;\;\;c \,=\,-\frac{8}{7}a$

$\text{Hence: }\:\vec w \;=\;\left\langle\,a,\,0,\,-\frac{8}{7}a\,\right\rangle$

$\text{Let }a\,=\,7:\;\vec w \;=\;\langle 7,\,0,\,-8\rangle$

 April 6th, 2015, 07:53 AM #3 Senior Member   Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4 Thanks! ! !

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