document.write( "Question 1117973: ) Consider the line L in R^3
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\n" ); document.write( "(a) Verify that the point (2, −1, 3) lies on L, but that (1, 1, 1) does not.
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Algebra.Com's Answer #733331 by ikleyn(52786) $\"\"$ $\"About$

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\n" ); document.write( "Consider the line L in R^3
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\n" ); document.write( "\n" ); document.write( "I edited your post to make my writing easier. In particular, I replaced $\"lambda\"$ by t, without changing the meaning.\r
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document.write( "a)  You are given the line \"parametrized by the parameter t\".\r\n" );
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document.write( "    In order to verify that the point (2,-1,3) lies on L,  you need to show that there is such \"t\" that\r\n" );
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document.write( "         2 = -t + 2,     (1)\r\n" );
document.write( "        -1 = 2t - 1,     (2)\r\n" );
document.write( "         3 =  t + 3.     (3)\r\n" );
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document.write( "    To do it,  from equation (1)  find  t = -2 + 2 = 0  and check that this value of t satisfies equations (2) and (3).\r\n" );
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document.write( "    You can easily check it on your own.\r\n" );
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document.write( "    So, this question is answered.\r\n" );
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document.write( "    Next question relates to the point (1,1,1),  and we need to show that THERE IS NO such t which satisfies the equations\r\n" );
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document.write( "         1 = -t + 2,     (4)\r\n" );
document.write( "         1 = 2t - 1,     (5)\r\n" );
document.write( "         1 =  t + 3.     (6)\r\n" );
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document.write( "    Indeed, from equation (4) t = -1 + 2 = 1.  The value of t= 1 still satisfies equation (5), but does not satisfy equation (6).\r\n" );
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document.write( "     So, this question is answered, too.\r\n" );
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document.write( "b)  The straight line L in   has the guiding vector (-1,2,1) comprised of the coefficients of its parametrization.\r\n" );
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document.write( "    The orthogonal plane to this vector has the form \r\n" );
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document.write( "        -1x + 2y + 1z = c,\r\n" );
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document.write( "     where \"c\" is some constant, whose value we determine using the condition that the point (2,-1,3) belongs to the plane.\r\n" );
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document.write( "     So, to find \"c\", we simply substitute the coordinates  x= 2, y= -1, z= 3 into equation (*) to get\r\n" );
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document.write( "        c = (-1)*2 + 2*(-1) + 1*3 = -2 -2 + 3 = -1.\r\n" );
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document.write( "    Thus your plane under the question b) has the equation\r\n" );
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document.write( "        -x + 2y + z = -1.\r\n" );
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