Last updated at June 3, 2021 by Teachoo

Transcript

Example 27 (Method 1) Find the equation of the plane that contains the point (1, โ1, 2) and is perpendicular to each of the planes 2x + 3y โ 2z = 5 and x + 2y โ 3z = 8. The equation of a plane passing through (๐ฅ_1, ๐ฆ_1, ๐ง_1) is given by A(x โ ๐_๐) + B (y โ ๐_๐) + C(z โ ๐_๐) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (1, โ1, 2) So, equation of plane is A(x โ1) + B (y + 1) + C(z โ 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 ๐ โ ร ๐ โ is perpendicular to both ๐ โ & ๐ โ So, required is normal is cross product of normal of planes 2x + 3y โ 2z = 5 and x + 2y โ 3z = 8. Required normal = |โ 8(๐ ฬ&๐ ฬ&๐ ฬ@2&3&โ2@1&2&โ3)| = ๐ ฬ (3(โ3) โ 2(โ2)) โ ๐ ฬ (2(โ3) โ 1(โ2)) + ๐ ฬ(2(2) โ 1(3)) = ๐ ฬ (โ9 + 4) โ ๐ ฬ (โ6 + 2) + ๐ ฬ(4 โ 3) = โ5๐ ฬ + 4๐ ฬ + ๐ ฬ Hence, direction ratios = โ5, 4, 1 โด A = โ5, B = 4, C = 1 Putting above values in (1), A(x โ1) + B (y + 1) + C(z โ 2) = 0 โ5(x โ 1) + 4 (y + 1) + 1 (z โ 2) = 0 โ5x + 5 + 4y + 4 + z โ 2 = 0 โ5x + 4y + z + 7 = 0 โ5x + 4y + z = โ7 โ(5x โ4y โ z) = โ7 5x โ 4y โ z = 7 Therefore, the equation of the required plane is 5x โ 4y โ z = 7. Example 27 (Method 2) Find the equation of the plane that contains the point (1, โ 1, 2) and is perpendicular to each of the planes 2x + 3y โ 2z = 5 and x + 2y โ 3z = 8. The equation of a plane passing through (๐ฅ_1, ๐ฆ_1, ๐ง_1) is given by A(x โ ๐_๐) + B (y โ ๐_๐) + C(z โ ๐_๐) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (1, โ1, 2) So, equation of plane is A(x โ1) + B (y + 1) + C(z โ 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 โ1) + B (y + 1) + C(z โ 2) = 0 is perpendicular to plane 2x + 3y โ 2z = 5 Hence, A ร 2 + B ร 3 + C ร (โ2) = 0 2A + 3B โ 2C = 0 Similarly, Given that plane A(x โ1) + B (y + 1) + C(z โ 2) = 0 is perpendicular to plane x + 2y โ 3z = 8 Two lines with direction ratios ๐_1, ๐_1, ๐_1 and ๐_2, ๐_2, ๐_2 are perpendicular if ๐_1 ๐_2 + ๐_1 ๐_2 + ๐_1 ๐_2 = 0 Hence, A ร 1 + B ร 2 + C ร (โ3) = 0 A + 2B โ 3C = 0 So, our equations are 2A + 3B โ2C = 0 A + 2B โ 3C = 0 Solving Two lines with direction ratios ๐_1, ๐_1, ๐_1 and ๐_2, ๐_2, ๐_2 are perpendicular if ๐_1 ๐_2 + ๐_1 ๐_2 + ๐_1 ๐_2 = 0 ๐ด/(โ9 โ (โ4)) = ๐ต/(โ2 โ (โ6)) = ๐ถ/(4 โ 3) ๐ด/(โ9 + 4) = ๐ต/(โ2 + 6) = ๐ถ/1 ๐ด/(โ5) = ๐ต/4 = ๐ถ/1 = k So, A = โ5k , B = 4k , C = k Putting above values in (1), A(x โ1) + B (y + 1) + C(z โ 2) = 0 โ5k(x โ 1) + 4k (y + 1) + k (z โ 2) = 0 k[โ5(x โ 1) + 4(y + 1) + (z โ 2)] = 0 โ5x + 5 + 4y + 4 + z โ 2 = 0 โ5x + 4y + z + 7 = 0 โ5x + 4y + z = โ7 โ(5x โ4y โ z) = โ7 5x โ 4y โ z = 7 Therefore, the equation of the required plane is 5x โ 4y โ z = 7.

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Example, 9 Deleted for CBSE Board 2022 Exams

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Example 22 Deleted for CBSE Board 2022 Exams

Example 23 Important Deleted for CBSE Board 2022 Exams

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Example 27 Important You are here

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Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.