document.write( "Question 961162: Two lines have equations \r
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\n" ); document.write( "\n" ); document.write( "Determine points P(on L1) and Q(on L2) such that PQ is perpendicular to both lines.
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Algebra.Com's Answer #587430 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
L1 :[x,y,z] = [1,3,2] + t[3,0,-1]
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document.write( "A direction vector of L1 is < 3,0,-1 >\r\n" );
document.write( "A direction vector of L2 is < 2,-1,3 > \r\n" );
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document.write( "A general point of L1 is\r\n" );
document.write( "P = (1+3t, 3, 2-t)\r\n" );
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document.write( "A general point of L2 is \r\n" );
document.write( "Q = (3+2s, 1-s, -2+3s) \r\n" );
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document.write( "The components of a vector from P to Q is\r\n" );
document.write( "found by subtracting their coordinates:\r\n" );
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document.write( "< -2+3t-2s,2+s,4-t-3s >\r\n" );
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document.write( "We want to make this vector be perpendicular\r\n" );
document.write( "to the direction vectors of L1 and L2 \r\n" );
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document.write( "< 3,0,-1 > and < 2,-1,3 >. \r\n" );
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document.write( "So we dot it with each of those and set\r\n" );
document.write( "each equal to 0: \r\n" );
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document.write( "< -2+3t-2s,2+s,4-t-3s > • < 3,0,-1 > = 0\r\n" );
document.write( "< -2+3t-2s,2+s,4-t-3s > • < 2,-1,3 > = 0\r\n" );
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document.write( "-6+9t-6s+0+0-4+t+3s = 0\r\n" );
document.write( "-4+6t-4s-2-s+12-3t-9s = 0\r\n" );
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document.write( "10t- 3s = 10\r\n" );
document.write( " 3t-14s =  6\r\n" );
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document.write( "Solve that system and get \r\n" );
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document.write( "t=122/131, s=-30/131\r\n" );
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document.write( "What god-awful fractions!  Hope I didn't\r\n" );
document.write( "make a mistake.\r\n" );
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document.write( "So to see what those points are we\r\n" );
document.write( "have to substitute those god-awful \r\n" );
document.write( "fractions in\r\n" );
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document.write( "P = (1+3t, 3, 2-t) = (497/131,3,140/131)\r\n" );
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document.write( "and\r\n" );
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document.write( "Q = (3+2s, 1-s, -2+3s) = (333/131,161/131,-352/131)\r\n" );
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document.write( "Edwin

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