Find the distance between the point and the plane. (2, 5, 3)
3x + 2y – 3z = –6
To find the distance between a point and a plane, you can use the formula:
Distance = |Ax + By + Cz + D| / √(A^2 + B^2 + C^2)
In this formula, (x, y, z) represents the coordinates of the point, and A, B, C, and D represent the coefficients of the equation of the plane.
In this case, the coordinates of the point are (2, 5, 3), and the coefficients of the equation of the plane are A = 3, B = 2, C = -3, and D = -6.
Substituting these values into the formula, we get:
Distance = |3(2) + 2(5) - 3(3) - 6| / √(3^2 + 2^2 + (-3)^2)
Simplifying the equation further, we have:
Distance = |6 + 10 - 9 - 6| / √(9 + 4 + 9)
Distance = |1| / √22
Distance = 1 / √22
Therefore, the distance between the point (2, 5, 3) and the plane 3x + 2y - 3z = -6 is 1 / √22, which is approximately 0.2134.