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What is the distance between the parallel planes 2x+y-3z=13 and 2x+y-3z=35?

  • Math -

    Pick any point on the first plane, say,
    (0,13,0)

    distance from that point to
    2x + y - 3z - 35 = 0

    = |2(0) + 13 - 3(0) - 35|/√(2^2 + 1^2 + (-3)^2)
    = 22/√14 or after rationalizing , 11√14/7

  • Math -

    simple. Find a point on the first plane. When x and z are zero, y=13.

    Now we find the distance between the point (0,13,0) and the second plane.

    d= (2*0+1*13-3*0 -35)/sqrt(4+1+9)
    d= 13/sqrt(14)

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