document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=389306>Question 389306</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If P(1, -2, 4) is reflected in the plane with equation 2x - 3y - 4z + 66 = 0, determine the coordinates of its image point, P'. Note that the plane 2x - 3y - 4z + 66 = 0 is the right bisector of the line joining  P(1, -2, 4) with its image.\r<BR>\n" );
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document.write( "Any help would be appreciated ^.^ </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #275855 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=275855\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First determine the equation of the line perpendicular to the given plane and passing through the point (1,-2,4).  Then get the intersection of the line with the given plane.  That intersection point is the midpoint of the segment connecting  (1,-2,4) to the unknown point (a,b,c).\r<BR>\n" );
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document.write( "The normal vector to the given plane is <2,-3,-4>.  The symmetric form of the line perpendicular to the given point is then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-1%29%2F2+=+%28y%2B2%29%2F%28-3%29+=+%28z-4%29%2F%28-4%29 ALIGN=MIDDLE  ALT=\"%28x-1%29%2F2+=+%28y%2B2%29%2F%28-3%29+=+%28z-4%29%2F%28-4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Solving for y and z in terms of x, then we get:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+%28-3x-1%29%2F2 ALIGN=MIDDLE  ALT=\"y+=+%28-3x-1%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and z = -2x+6.  \r<BR>\n" );
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document.write( "Putting these two equations into 2x - 3y - 4z + 66 = 0, we get<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x-3%28%28-3x-1%29%2F2%29+-+4%28-2x%2B6%29+%2B+66+=+2x+%2B+9x%2F2+%2B+3%2F2+%2B+8x-24+%2B+66+=+0 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=29x%2F2+=+-87%2F2 ALIGN=MIDDLE  ALT=\"29x%2F2+=+-87%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > after simplification.<BR>\n" );
document.write( "Hence x = -3, y = 4, and z = 12, after substituting back into <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+%28-3x-1%29%2F2 ALIGN=MIDDLE  ALT=\"y+=+%28-3x-1%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > and z = -2x+6.  Thus the intersection point of the normal line and the plane is (-3,4,12). <BR>\n" );
document.write( "To find the reflected point (a,b,c), we use the midpoint formula:  (Recall the given point is (1,-2,4))\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2Ba%29%2F2+=+-3 ALIGN=MIDDLE  ALT=\"%281%2Ba%29%2F2+=+-3\"   DISABLED_style= \"max-width:90%; height:auto;\" > ----> a = -7<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-2+%2B+b%29%2F2+=+4 ALIGN=MIDDLE  ALT=\"%28-2+%2B+b%29%2F2+=+4\"   DISABLED_style= \"max-width:90%; height:auto;\" > ----> b = 10<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%284%2Bc%29%2F2+=+12 ALIGN=MIDDLE  ALT=\"%284%2Bc%29%2F2+=+12\"   DISABLED_style= \"max-width:90%; height:auto;\" > ------> c = 20.\r<BR>\n" );
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document.write( "Therefore the reflected point is (-7,10,20).<BR>\n" );
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