Check sibling questions

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.