document.write( "Question 54307This question is from textbook Linear Algebra
\n" ); document.write( ": find the distance from point Q to line l (matriks)\r
\n" ); document.write( "\n" ); document.write( "Q=(0, 1, 0)\r
\n" ); document.write( "\n" ); document.write( "l= [x y z] = [1 1 1] + t[-2 0 3] \n" ); document.write( "

Algebra.Com's Answer #36559 by Edwin McCravy(20054) $\"\"$ $\"About$

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find the distance from point Q to line l (matrix)\r\n" );
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document.write( "Q=(0, 1, 0)\r\n" );
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document.write( "line = < x y z > = < 1 1 1 > + t< -2 0 3 >\r\n" );
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document.write( "Formula:\r\n" );
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document.write( "     ||PQ × u||\r\n" );
document.write( "D = ------------\r\n" );
document.write( "       ||u||\r\n" );
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document.write( "where u is the direction vector for the line and\r\n" );
document.write( "P is a point on the line\r\n" );
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document.write( "Using the direction numbers -2, 0, and 4, you know\r\n" );
document.write( "that the direction vector for the line is\r\n" );
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document.write( "u = < -2, 0, 4 >\r\n" );
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document.write( "To find a point on the line, let t = 0 and obtain\r\n" );
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document.write( "P = (1, 1, 1)\r\n" );
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document.write( "Thus, since Q = (0, 1, 0)\r\n" );
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document.write( "PQ = < 0-1, 1-1, 0-1 > = < -1,0,-1 >\r\n" );
document.write( "                          __\r\n" );
document.write( "Now get the cross product PQ × u\r\n" );
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document.write( "__       | i  j  k| \r\n" );
document.write( "PQ × u = |-1  0 -1| = i(0+0) - j(-4-2) + k(0+0) = -6j\r\n" );
document.write( "         |-2  0  4|\r\n" );
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document.write( "||PQ × u|| = ||-6j|| = 6\r\n" );
document.write( "         ____________    ______    __     _\r\n" );
document.write( "||u|| = Ö(-2)²+0²+ 4² = Ö4+0+16 = Ö20 = 2Ö5\r\n" );
document.write( "       __                            _\r\n" );
document.write( "     ||PQ × u||      6       3     3Ö5\r\n" );
document.write( "D = ------------ = ----- = ---- = -----\r\n" );
document.write( "       ||u||        2Ö5     Ö5      5\r\n" );
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document.write( "Edwin

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