Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 8

Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 9
Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 10

Something went wrong!

The video couldn't load due to a technical hiccup.
But don't worry — our team is already on it, and we're working hard to get it back up ASAP.

Thanks for bearing with us!

Share on WhatsApp

Transcript

Question 1 In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (d) 5y + 8 = 0 For plane ax + by + cz = d Direction ratios of normal = a, b, c Direction cosines : l = 𝑎/√(𝑎^2 + 𝑏^2 +〖 𝑐〗^2 ) , m = 𝑏/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) , n = 𝑐/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) Distance from origin = 𝑑/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) Given, equation of the plane is 5y + 8 = 0 5y = −8 −5y = 8 0x − 5y + 0z = 8 0x − 5y + 0z = 8 Comparing with ax + by + cz = d a = 0, b = –5, c = 0 & d = 8 & √(𝑎^2+𝑏^2+𝑐^2 ) = √(0^2 + 〖(−5)〗^2 + 0^2 ) = √25 = 5 Direction cosines of the normal to the plane are l = 𝑎/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) , m = 𝑏/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) , n = 𝑐/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) l = 0/5, m = (−5)/5, n = ( 0)/5 ∴ Direction cosines of the normal to the plane are = (0, –1, 0) And, Distance form the origin = 𝑑/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) = 𝟖/𝟓

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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo