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Use normal vectors to determine the intersection, if any, for each of the following groups of three planes. Give a geometric interpretation in each case and the number of solutions for the corresponding linear system of equations. If the planes intersect in a line, determine a vector equation of the line. If the planes intersect in a point, determine the coordinates of the point.

x − y + 3z = 4
x + y + 2z = 2
3x + y + 7z = 9

  • Calculus - ,

    adding #1 and #2 I got 2x + 5z = 6
    adding #1 and #3 I got 4x + 10z = 13

    You should realize that we cannot solve this any further, since both x and z would drop out

    so lets find the line of intersection of plane #1 and #2

    from 2x + 5z = 6
    let z=2, then x = -2 and from #2, y = 0
    let z=4, then x = -7 and from #2 y = 1

    so we have two points on the line of intersection of the first two planes, namely
    (-2,0,2) and (-7,1,4) from which we can get a direction vector (5,-1,-2)

    and parametric equations of that line could be
    x = -2+5t
    y = -t
    z = 2-2t

    subbing that into the third equation 3x + y + 7z = 0 we get
    -6 + 15t - t + 14 - 14t = 9
    8=9 which is a contradiction.

    Had this been a TRUE statement then all 3 planes would have intersected in the same line, like the spine of an open book

    but the statement was FALSE, so the 3 planes must intersect in 3 different but parallel lines,

    to find the other two line equations follow my procedure above for the other two possible pairs of planes.

    As a check they should all have the same direction numbers.

    BTW, this question is the hardest of the possible types of intersection of 3 planes.

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