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Example 27 - Find equation of plane that contains the point (1, -1, 2)

Example 27 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 27 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

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Example 27 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

Example 27 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5
Example 27 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6
Example 27 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7
Example 27 - Chapter 11 Class 12 Three Dimensional Geometry - Part 8

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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 (π‘₯_1, 𝑦_1, 𝑧_1) is given by A(x βˆ’ 𝒙_𝟏) + B (y βˆ’ π’š_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, βˆ’1, 2) So, equation of plane is A(x βˆ’1) + B (y + 1) + C(z βˆ’ 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 normal of both planes. We know that π‘Ž βƒ— Γ— 𝑏 βƒ— is perpendicular to both π‘Ž βƒ— & 𝑏 βƒ— So, required is normal is cross product of normal of planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Required normal = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@2&3&βˆ’2@1&2&βˆ’3)| = 𝑖 Μ‚ (3(–3) – 2(–2)) – 𝑗 Μ‚ (2(–3) – 1(–2)) + π‘˜ Μ‚(2(2) – 1(3)) = 𝑖 Μ‚ (–9 + 4) – 𝑗 Μ‚ (–6 + 2) + π‘˜ Μ‚(4 – 3) = –5𝑖 Μ‚ + 4𝑗 Μ‚ + π‘˜ Μ‚ Hence, direction ratios = –5, 4, 1 ∴ A = –5, B = 4, C = 1 Putting above values in (1), A(x βˆ’1) + B (y + 1) + C(z βˆ’ 2) = 0 βˆ’5(x βˆ’ 1) + 4 (y + 1) + 1 (z βˆ’ 2) = 0 βˆ’5x + 5 + 4y + 4 + z βˆ’ 2 = 0 βˆ’5x + 4y + z + 7 = 0 βˆ’5x + 4y + z = βˆ’7 βˆ’(5x βˆ’4y βˆ’ z) = βˆ’7 5x βˆ’ 4y βˆ’ z = 7 Therefore, the equation of the required plane is 5x βˆ’ 4y βˆ’ 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 (π‘₯_1, 𝑦_1, 𝑧_1) is given by A(x βˆ’ 𝒙_𝟏) + B (y βˆ’ π’š_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, βˆ’1, 2) So, equation of plane is A(x βˆ’1) + B (y + 1) + C(z βˆ’ 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 βˆ’1) + B (y + 1) + C(z βˆ’ 2) = 0 is perpendicular to plane 2x + 3y – 2z = 5 Hence, A Γ— 2 + B Γ— 3 + C Γ— (–2) = 0 2A + 3B βˆ’ 2C = 0 Similarly, Given that plane A(x βˆ’1) + B (y + 1) + C(z βˆ’ 2) = 0 is perpendicular to plane x + 2y – 3z = 8 Two lines with direction ratios π‘Ž_1, 𝑏_1, 𝑐_1 and π‘Ž_2, 𝑏_2, 𝑐_2 are perpendicular if π‘Ž_1 π‘Ž_2 + 𝑏_1 𝑏_2 + 𝑐_1 𝑐_2 = 0 Hence, A Γ— 1 + B Γ— 2 + C Γ— (–3) = 0 A + 2B βˆ’ 3C = 0 So, our equations are 2A + 3B βˆ’2C = 0 A + 2B βˆ’ 3C = 0 Solving Two lines with direction ratios π‘Ž_1, 𝑏_1, 𝑐_1 and π‘Ž_2, 𝑏_2, 𝑐_2 are perpendicular if π‘Ž_1 π‘Ž_2 + 𝑏_1 𝑏_2 + 𝑐_1 𝑐_2 = 0 𝐴/(βˆ’9 βˆ’ (βˆ’4)) = 𝐡/(βˆ’2 βˆ’ (βˆ’6)) = 𝐢/(4 βˆ’ 3) 𝐴/(βˆ’9 + 4) = 𝐡/(βˆ’2 + 6) = 𝐢/1 𝐴/(βˆ’5) = 𝐡/4 = 𝐢/1 = k So, A = –5k , B = 4k , C = k Putting above values in (1), A(x βˆ’1) + B (y + 1) + C(z βˆ’ 2) = 0 βˆ’5k(x βˆ’ 1) + 4k (y + 1) + k (z βˆ’ 2) = 0 k[βˆ’5(x βˆ’ 1) + 4(y + 1) + (z βˆ’ 2)] = 0 βˆ’5x + 5 + 4y + 4 + z βˆ’ 2 = 0 βˆ’5x + 4y + z + 7 = 0 βˆ’5x + 4y + z = βˆ’7 βˆ’(5x βˆ’4y βˆ’ z) = βˆ’7 5x βˆ’ 4y βˆ’ z = 7 Therefore, the equation of the required plane is 5x βˆ’ 4y βˆ’ z = 7.

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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.