Check sibling questions

Example 24 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Last updated at Dec. 24, 2019 by

Example 24 - Find distance of point (2, 5, -3) from plane - Examples

Example 24 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

This video is only available for Teachoo black users

Subscribe Now

Solve all your doubts with Teachoo Black (new monthly pack available now!)

Join Teachoo Black

Transcript

Example 24 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﷮ ﷮ 6﷮2﷯ + −3﷯﷮2﷯ + 2﷮2﷯﷯﷯﷯ = 2 × 6﷯ + 5 × −3﷯ + −3 × 2﷯ − 4﷮ ﷮36 + 9 + 4﷯﷯﷯ = 12 − 15 − 6 − 4﷮ ﷮49﷯﷯﷯ = −13﷮7﷯﷯ = 𝟏𝟑﷮𝟕﷯

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.