Last updated at March 11, 2017 by Teachoo

Transcript

Example 27 (Method 1) Find the equation of the plane that contains the point (1, – 1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. The equation of a plane passing through ( 𝑥1, 𝑦1, 𝑧1) is given by A(x − 𝒙𝟏) + B (y − 𝒚𝟏) + C(z – 𝒛𝟏) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (1, −1, 2) So, equation of plane is A(x −1) + B (y + 1) + C(z − 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C Also, the plane is perpendicular to the given two planes, So, their normal to plane would be perpendicular to normals of both planes. We know that 𝑎 × 𝑏 is perpendicular to both 𝑎 & 𝑏 So, required is normal is cross product of normals of planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Required normal = 𝑖 𝑗 𝑘23−212−3 = 𝑖 (3(–3) – 2(–2)) – 𝑗 (2(–3) – 1(–2)) + 𝑘(2(2) – 1(3)) = 𝑖 (–9 + 4) – 𝑗 (–6 + 2) + 𝑘(4 – 3) = –5 𝑖 + 4 𝑗 + 𝑘 Hence, direction ratios = –5, 4, 1 ∴ A = –5, B = 4, C = 1 Putting above values in (1), A(x −1) + B (y + 1) + C(z − 2) = 0 −5(x − 1) + 4 (y + 1) + 1 (z − 2) = 0 −5x + 5 + 4y + 4 + z − 2 = 0 −5x + 4y + z + 7 = 0 −5x + 4y + z = −7 −(5x −4y − z) = −7 5x − 4y − z = 7 Therefore, the equation of the required plane is 5x − 4y − z = 7. Example 27 (Method 2) Find the equation of the plane that contains the point (1, – 1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. The equation of a plane passing through ( 𝑥1, 𝑦1, 𝑧1) is given by A(x − 𝒙𝟏) + B (y − 𝒚𝟏) + C(z – 𝒛𝟏) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (1, −1, 2) So, equation of plane is A(x −1) + B (y + 1) + C(z − 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C Also, the plane is perpendicular to the given two planes. Now, it is given that plane A(x −1) + B (y + 1) + C(z − 2) = 0 is perpendicular to plane 2x + 3y – 2z = 5 Hence, A × 2 + B × 3 + C × (–2) = 0 2A + 3B − 2C = 0 Similarly, Given that plane A(x −1) + B (y + 1) + C(z − 2) = 0 is perpendicular to plane x + 2y – 3z = 8 Hence, A × 1 + B × 2 + C × (–3) = 0 A + 2B − 3C = 0 So, our equations are 2A + 3B −2C = 0 …(2) A + 2B − 3C = 0 …(3) Solving 𝐴−9 − (−4) = 𝐵−2 − (−6) = 𝐶4 − 3 𝐴−9 + 4 = 𝐵−2 + 6 = 𝐶1 𝐴−5 = 𝐵4 = 𝐶1 = k So, A = –5k , B = 4k , C = k Putting above values in (1), A(x −1) + B (y + 1) + C(z − 2) = 0 −5k(x − 1) + 4k (y + 1) + k (z − 2) = 0 k[−5(x − 1) + 4(y + 1) + (z − 2)] = 0 −5x + 5 + 4y + 4 + z − 2 = 0 −5x + 4y + z + 7 = 0 −5x + 4y + z = −7 −(5x −4y − z) = −7 5x − 4y − z = 7 Therefore, the equation of the required plane is 5x − 4y − z = 7.

Example 1

Example, 2

Example, 3 Important

Example, 4

Example, 5

Example, 6 Important

Example, 7

Example 8

Example, 9 Important

Example 10

Example 11

Example 12 Important

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18

Example 19

Example 20 Important

Example 21 Important

Example 22

Example 23 Important

Example 24 Important

Example, 25 Important

Example 26

Example 27 Important You are here

Example 28

Example 29 Important

Example 30 Important

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .