document.write( "Question 892544: Determining an orthogonal basis for W = {(x, y, z); x - 2y + z = 0}. \n" ); document.write( "

Algebra.Com's Answer #540592 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
Choose any vector in W.
\n" ); document.write( "Set $\"x=0\"$ , then $\"-2y%2Bz=0\"$
\n" ); document.write( " $\"2y=z\"$
\n" ); document.write( "If $\"y=1\"$ , then $\"z=2\"$
\n" ); document.write( "(0,1,2)
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "Now use the dot product to find a perpendicular vector to this vector.
\n" ); document.write( " $\"0%2Aa%2B1%2Ab%2B2%2Ac=0\"$
\n" ); document.write( "Let $\"a=1\"$
\n" ); document.write( "then $\"b=-2c\"$
\n" ); document.write( "Let $\"c=1\"$ , then $\"b=-2\"$
\n" ); document.write( "(1,-2,1)
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".\r
\n" ); document.write( "\n" ); document.write( "Now take the cross product of those two vectors to find a mutually perpendicular vector to these two.
\n" ); document.write( "(0,1,2)X(1,-2,1)=(5,2,-1)
\n" ); document.write( "So then,
\n" ); document.write( "(0,1,2), (1,-2,1), and (5,2,-1) form an orthogonal basis of W.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );