1. Class 12
2. Important Question for exams Class 12
3. Chapter 11 Class 12 Three Dimensional Geometry

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 ( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) is given by A(x − 𝒙﷮𝟏﷯) + B (y − 𝒚﷮𝟏﷯) + C(z – 𝒛﷮𝟏﷯) = 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﷯﷮ 𝑘﷯﷮2﷮3﷮−2﷮1﷮2﷮−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 ( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) is given by A(x − 𝒙﷮𝟏﷯) + B (y − 𝒚﷮𝟏﷯) + C(z – 𝒛﷮𝟏﷯) = 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)﷯ = 𝐶﷮4 − 3﷯ 𝐴﷮−9 + 4﷯ = 𝐵﷮−2 + 6﷯ = 𝐶﷮1﷯ 𝐴﷮−5﷯ = 𝐵﷮4﷯ = 𝐶﷮1﷯ = 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.

Chapter 11 Class 12 Three Dimensional Geometry

Class 12
Important Question for exams Class 12

About the Author

Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.