document.write( "Question 1010756: consider the following scalar equation of a plane. x+3y+z=0\r
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document.write( "Find the orthogonal complement of the vector <1,2,1>. Also find the point on the plane which is closest to (3,4,1) \n" );
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Algebra.Com's Answer #626827 by Edwin McCravy(20060)![]() ![]() You can put this solution on YOUR website! consider the following scalar equation of a plane. x+3y+z=0\r \n" ); document.write( "\n" ); document.write( "Find the orthogonal complement of the vector <1,2,1>. \n" ); document.write( " \r\n" ); 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" ); document.write( "\r\n" ); document.write( "Vectors in or parallel to the plane are vectors < x,y,z > such that \r\n" ); document.write( "\r\n" ); document.write( "x+2y+z = 0\r\n" ); document.write( "\r\n" ); document.write( "Vectors perpendicular to <1,3,1> are < x,y,z > such that\r\n" ); document.write( "\r\n" ); document.write( "<1,3,1>∙< x,y,z > = x+3y+z = 0\r\n" ); document.write( "\r\n" ); document.write( "So we have this under-determined system\r\n" ); document.write( "\r\n" ); document.write( " \n" ); document.write( "Also find the point on the plane which is closest to (3,4,1). \n" ); document.write( " \r\n" ); 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" ); document.write( " \r\n" ); document.write( "(x1, y1, z1) to the plane\r\n" ); document.write( "\r\n" ); document.write( "ax + by + cz + d = 0, which is:\r\n" ); document.write( "\r\n" ); document.write( "is\n" ); document.write( " |