document.write( "Question 1201853: this is from my 'Cartesian Equation of a Plane' lesson.\r
\n" ); document.write( "\n" ); document.write( "a. Determine the Cartesian equation of the plane that contains the three points A(-2,3,1), B(3,4,5) and C(1,1,0).\r
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\n" ); document.write( "\n" ); document.write( "b. Determine the Cartesian equation of the plane containing the point P(-1,1,0) and perpendicular to the line between the points A(1,2,1) and B(3,-2,0). \r
\n" ); document.write( "\n" ); document.write( "thank you.
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Algebra.Com's Answer #836451 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Answers:
\n" ); document.write( "(a) 7x+17y-13z = 24
\n" ); document.write( "(b) 2x-4y-z = -6
\n" ); document.write( "Other answers are possible.\r
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\n" ); document.write( "\n" ); document.write( "Explanation for part (a)\r
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\n" ); document.write( "\n" ); document.write( "We have these three points
\n" ); document.write( "A(-2,3,1)
\n" ); document.write( "B(3,4,5)
\n" ); document.write( "C(1,1,0)\r
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\n" ); document.write( "\n" ); document.write( "Subtract the coordinates of B and A to find the vector that points from A to B.
\n" ); document.write( "B-A = < 3,4,5 > - < -2,3,1 >
\n" ); document.write( "B-A = < 3,4,5 > + < 2,-3,-1 >
\n" ); document.write( "B-A = < 3+2,4-3,5-1 >
\n" ); document.write( "B-A = < 5,1,4 >
\n" ); document.write( "This says to go from A to B, we do three things:
  • Move 5 units along the positive x axis.
  • Move 1 unit along the positive y axis.
  • Move 4 units along the positive z axis.
Therefore, vector AB is < 5,1,4 >
\n" ); document.write( "The order in vector naming is important. \"Vector AB\" means we start at A and point to B.
\n" ); document.write( "While \"vector BA\" means we start at B and point at A.\r
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\n" ); document.write( "\n" ); document.write( "Repeat similar steps to find vector AC = < 3,-2,-1 >\r
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\n" ); document.write( "\n" ); document.write( "We found that
\n" ); document.write( "vector AB = < 5,1,4 >
\n" ); document.write( "vector AC = < 3,-2,-1 >\r
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\n" ); document.write( "\n" ); document.write( "Next, take the cross product of these two vectors.
\n" ); document.write( "This will construct a vector perpendicular to both AB and AC,
\n" ); document.write( "Think of this new vector as a vertical pole out of the flat horizontal ground. Vectors AB and AC are entirely on the flat ground.
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\n" ); document.write( "\n" ); document.write( "The cross product of vectors AB and AC is < 7, 17, -13 > as discussed in this lesson here
\n" ); document.write( "https://www.algebra.com/algebra/homework/Vectors/cross-product.lesson
\n" ); document.write( "This represents the normal vector to the plane.
\n" ); document.write( "We can think of the normal vector being of the form < a,b,c >
\n" ); document.write( "In this case < a,b,c > = < 7,17,-13 >
\n" ); document.write( "The normal vector tells us how to tilt the plane.\r
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\n" ); document.write( "\n" ); document.write( "One template equation for a plane is
\n" ); document.write( "a(x - p) + b(y - q) + c(z - r) = 0
\n" ); document.write( "where
\n" ); document.write( "a,b,c = coordinates of the normal vector we just found
\n" ); document.write( "p,q,r = coordinates of a point that is on the plane\r
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\n" ); document.write( "\n" ); document.write( "We have 3 choices for what we pick for p,q,r
\n" ); document.write( "I'll go for (p,q,r) = (-2,3,1) which is the location of point A.
\n" ); document.write( "You could pick the coordinates of B or C.\r
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\n" ); document.write( "\n" ); document.write( "So,
\n" ); document.write( "a(x-p) + b(y-q) + c(z - r) = 0
\n" ); document.write( "7(x-p) + 17(y-q) - 13(z - r) = 0 ... plugging in coordinates from normal vector
\n" ); document.write( "7(x-(-2)) + 17(y-3) - 13(z - 1) = 0 ... plugging in coordinates from point A in the plane
\n" ); document.write( "7(x+2) + 17(y-3) - 13(z - 1) = 0
\n" ); document.write( "7x+14 + 17y-51 - 13z + 13 = 0
\n" ); document.write( "7x+17y-13z + 14-51+13 = 0
\n" ); document.write( "7x+17y-13z - 24 = 0
\n" ); document.write( "7x+17y-13z = 24
\n" ); document.write( "That is one possible answer for part (a).
\n" ); document.write( "Other answers are possible because we could scale the equation up or down by some factor.\r
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\n" ); document.write( "\n" ); document.write( "Explanation for part (b)\r
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\n" ); document.write( "\n" ); document.write( "Let's find vector AB
\n" ); document.write( "vector AB = B-A
\n" ); document.write( "vector AB = < 3,-2,0 > - < 1,2,1 >
\n" ); document.write( "vector AB = < 3-1,-2-2,0-1 >
\n" ); document.write( "vector AB = < 2,-4,-1 >
\n" ); document.write( "The direction vector of line AB is < 2,-4,-1 >
\n" ); document.write( "Because this line is perpendicular to the plane we want, vector AB is a normal vector of this plane.\r
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\n" ); document.write( "\n" ); document.write( "normal vector = < a,b,c > = < 2,-4,-1 >
\n" ); document.write( "point on plane = (p,q,r) = (-1,1,0)\r
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\n" ); document.write( "\n" ); document.write( "Equation of the plane in cartesian form
\n" ); document.write( "a(x-p) + b(y-q) + c(z - r) = 0
\n" ); document.write( "2(x - p) - 4(y-q) - 1(z - r) = 0 ... plugging in coordinates from normal vector
\n" ); document.write( "2(x - (-1)) - 4(y - 1) - 1(z - 0) = 0 ... plugging in coordinates from point
\n" ); document.write( "2(x+1) - 4(y-1) - z = 0
\n" ); document.write( "2x + 2 - 4y + 4 - z = 0
\n" ); document.write( "2x-4y-z+6 = 0
\n" ); document.write( "2x+4y-z = -6\r
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\n" ); document.write( "\n" ); document.write( "Further Reading
\n" ); document.write( "https://www.whitman.edu/mathematics/calculus_online/section12.05.html
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