points and planes in 3d?

Question

The point A is the foot of the perpendicular from the point (1,1,9) to the plane 2x+y-z=6. Find the coordinates of A.

Can anyone explain how this is done? Thanks in advance!

Answer 1

Find the point of intersection A, of the perpendicular line thru the point P(1, 1, 9) and the plane 2x + y - z = 6.

2x + y - z - 6 = 0

The normal vector n, of the plane is also the directional vector of the perpendicular line thru P. The vector can be obtained directly from the coefficients of the variables in the equation of the plane.

n = <2, 1, -1>

With the directional vector n, and a point on the line P, we can write the equation of the perpendicular line.

L(t) = P + tn
L(t) = P(1, 1, 9) + t<2, 1, -1>
L(t) = <1 + 2t, 1 + t, 9 - t>
where t is a scalar ranging over the real numbers

The values of the variables x, y, and z of the line and the plane are equal at the point of intersection. Take the values of x, y, and z from the line and substitute it into the equation of the plane.

2x + y - z - 6 = 0
2(1 + 2t) + 1(1 + t) - 1(9 - t) - 6 = 0
2 + 4t + 1 + t - 9 + t - 6 = 0
6t - 12 = 0
6t = 12
t = 2

Solve for x, y, and z at t = 2.

x = 1 + 2t = 1 + 2*2 = 5
y = 1 + t = 1 + 2 = 3
z = 9 - t = 9 - 2 = 7

The point of intersection is A(5, 3, 7).

Answer 2

sure, they might style a airplane. Technically speaking, the ranges of freedom (i.e. the measurement of the manifold) is often decreased via a million whilst intersecting any 2 manifolds of equivalent measurement (except the two manifolds are same).