Find the distance from the point P(5, 3, 2) to the plane x + 2y + z = 6.
for a given point (p,q,r) and plane Ax + By + Cz + D = 0
the distance between the point and the plane is
|Ap + Bq + Cr + D| / √(A^2 + B^2 + C^2)
so you have (5,3,2) as the point and x + 2y + z - 6 = 0 as the plane
go for it
To find the distance from a point to a plane, you can use the formula:
distance = |ax + by + cz + d| / sqrt(a^2 + b^2 + c^2)
where (a, b, c) is the normal vector to the plane and (x, y, z) are the coordinates of the point, and d is the constant term in the equation of the plane.
In this case, the plane is given by the equation x + 2y + z = 6, so we can see that the normal vector to the plane is (1, 2, 1).
The point P is given by the coordinates (5, 3, 2).
We can substitute these values into the formula:
distance = |1 * 5 + 2 * 3 + 1 * 2 + 6| / sqrt(1^2 + 2^2 + 1^2)
= |5 + 6 + 2 + 6| / sqrt(1 + 4 + 1)
= |19| / sqrt(6)
= 19 / sqrt(6)
So, the distance from the point P(5, 3, 2) to the plane x + 2y + z = 6 is 19 / sqrt(6) units.