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Find the perpendicular distance from the point `(1,-3,4)` to the plane `3x-4y+12z - 1 = 0`. |
Answer» Perpendicular distance from point `(1,-3,4)` to the plane `3x-4y+12z-1 = 0` is `|(3(1)-4(-3)+12(4)-1)/(sqrt(3^(2)+(-4)^(2)+12^(2)))|` `= |(3+12+48-1)/(13)|=62/13` units. |
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