Olympiad Maths Prep

Track / Stage 3 / 63 of 260 #63 of 2000

Problem 63

AMC 10/12, early questions
Geometry Difficulty 3.2 Find the answer

In the rectangular coordinate system OxyzO-xyz, a normal vector of the plane OABOAB is n=(2,2,1)\overrightarrow{n}=(2,-2,1). Given point P(1,3,2)P(-1,3,2), the distance dd from point PP to the plane OABOAB is equal to _____.

Official solution

A normal vector of the plane OABOAB is n=(2,2,1)\overrightarrow{n}=(2,-2,1), and the given point is P(1,3,2)P(-1,3,2).

The distance dd from point PP to the plane OABOAB is given by the formula:

d=nOPn=26+222+(2)2+12=2d=\frac{|\overrightarrow{n}\cdot\overrightarrow{OP}|}{|\overrightarrow{n}|}=\frac{|-2-6+2|}{\sqrt{2^2+(-2)^2+1^2}}=2

Therefore, the answer is 2\boxed{2}.

This problem can be solved directly by using the formula for the distance between a point and a plane.

This question tests the knowledge of how to calculate distances between points, lines, and planes in space, as well as the application of formulas, and it is a basic question.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.