






Last updated at Feb. 4, 2020 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.
Examples
Example, 2 Important
Example, 3
Example, 4 Important
Example, 5 Important
Example, 6 Important
Example, 7
Example 8
Example, 9 Deleted for CBSE Board 2021 Exams only
Example 10 Deleted for CBSE Board 2021 Exams only
Example 11
Example 12 Important
Example 13 Important
Example 14
Example 15
Example 16 Important
Example 17
Example 18
Example 19 Important
Example 20 Important
Example 21 Important
Example 22 Deleted for CBSE Board 2021 Exams only
Example 23 Important Deleted for CBSE Board 2021 Exams only
Example 24
Example, 25 Important
Example 26
Example 27 Important You are here
Example 28 Important
Example 29 Important
Example 30 Important
About the Author