![Example 24 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2](https://d1avenlh0i1xmr.cloudfront.net/0f418648-7c92-467b-a4f1-fda4d46096e0/slide17.jpg)
Examples
Last updated at April 16, 2024 by Teachoo
Question 14 Find the distance of a point (2, 5, –3) from the plane 𝑟 . (6 𝑖 – 3 𝑗 + 2 𝑘) = 4 The distance of a point with position vector 𝑎 from the plane 𝑟. 𝑛 = d, where 𝑛 is the normal to the plane is 𝒂. 𝒏 − 𝒅 𝒏 Given, the point is (2, 5, −3) So, 𝑎 = 2 𝑖 + 5 𝑗 − 3 𝑘 The equation of plane is 𝑟.(6 𝑖 − 3 𝑗 + 2 𝑘) = 4 Comparing with 𝑟. 𝑛 = d, 𝑛 = 6 𝑖 − 3 𝑗 + 2 𝑘 & d = 4 Distance of point from plane = 𝑎. 𝑛 − 𝑑 𝑛 = 2 𝑖 + 5 𝑗 − 3 𝑘. 6 𝑖 − 3 𝑗 + 2 𝑘 − 4 62 + −32 + 22 = 2 × 6 + 5 × −3 + −3 × 2 − 4 36 + 9 + 4 = 12 − 15 − 6 − 4 49 = −137 = 𝟏𝟑𝟕