document.write( "Question 1010756: consider the following scalar equation of a plane. x+3y+z=0\r
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Algebra.Com's Answer #626827 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
consider the following scalar equation of a plane. x+3y+z=0\r
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document.write( "We want to find the set of vectors parallel to the plane that are\r\n" );
document.write( "perpendicular to the vector <1,2,1>.\r\n" );
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document.write( "Vectors in or parallel to the plane are vectors < x,y,z > such that \r\n" );
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document.write( "x+2y+z = 0\r\n" );
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document.write( "Vectors perpendicular to <1,3,1> are < x,y,z > such that\r\n" );
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document.write( "<1,3,1>∙< x,y,z > = x+3y+z = 0\r\n" );
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document.write( "So we have this under-determined system\r\n" );
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document.write( "\"system%28x%2B2y%2Bz=0%2Cx%2B3y%2Bz=0%29\"\r\n" );
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document.write( "Subtracting them we get y = 0\r\n" );
document.write( "Substituting we get x+2(0)+z = 0, or x+z=0, or z=-x\r\n" );
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document.write( "So the orthogonal complement is the set of vectors of the form\r\n" );
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document.write( "< k,0,-k >\r\n" );
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\n" ); document.write( "Also find the point on the plane which is closest to (3,4,1).
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document.write( "That is the perpendicular distance from the point. We use the\r\n" );
document.write( "formula for the perpendicular distance from the point\r\n" );
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document.write( "(x1, y1, z1) to the plane\r\n" );
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document.write( "ax + by + cz + d = 0, which is:\r\n" );
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document.write( "is \"D\"\"%22%22=%22%22\"\"abs%28ax%5B1%5D%2Bbx%5B1%5D%2Bcx%5B1%5D%2Bd%29%2Fsqrt%28a%5E2%2Bb%5E2%2Bc%5E2%29\"\r\n" );
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document.write( "So in our case (x1, y1, z1) = (3,4,1),\r\n" );
document.write( "a=1, b=3, c=1, d=0\r\n" );
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document.write( "\"D\"\"%22%22=%22%22\"\r\n" );
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document.write( "\"D\"\"%22%22=%22%22\"\"abs%283%2B12%2B1%29%2Fsqrt%281%2B3%2B1%29\"\r\n" );
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document.write( "\"D\"\"%22%22=%22%22\"\"16%2Fsqrt%285%29\"\r\n" );
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document.write( "Edwin
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