, p is a point outside α, a is any point inside α, and the distance from p to plane α→ is d, then d = | v pa |/| v |.

Analysis: A plane α is known.

V=(x 1, y2, z 1), p is the point out of plane, and the vector AP=(x2, y2, z2).

∵cos & lt； Vector v, vector PA>= | vector v vector PA |/| vector v vector PA |

because

That is, the distance from d to plane α is that the vector is in the plane.

Projected onto

∴d=| vector v vector PA |/| vector PA |

2.

Distance: Let the straight line n be equal to

A and B are vertical vectors, A and B are arbitrary points on A and B respectively, and D is the distance between A and B, then D = | AB N |/| N | Analysis: This formula is essentially the same as the above formula for the distance from point to plane.

∫n is related to the following factors

A vector in which both a and b are perpendicular.

Let the straight line a∈ face α and the straight line b// face α.

∴ Vector n is surface α.

A is any point on the straight line A, and ∴A is a point on the plane α.

Is any point on a straight line B out of plane.

The distance from b to surface α is equal to 2.

Apply the above formula.

D=| Vector AB Vector n|/| Vector n| I hope it will help you forget to adopt it.