Miscellaneous

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Question 9 (Method 1) Find the equation of the plane passing through the point (โ1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. The equation of a plane passing through (๐ฅ_1, ๐ฆ_1, ๐ง_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 (โ1, 3, 2) So, equation of plane is A(x + 1) + B (y โ 3) + C(z โ 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C The equation of a plane passing through (๐ฅ_1, ๐ฆ_1, ๐ง_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 (โ1, 3, 2) So, equation of plane is A(x + 1) + B (y โ 3) + 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 normals of planes x + 2y + 3z = 5 and 3x + 3y + z = 0 Required normal = |โ 8(๐ ฬ&๐ ฬ&๐ ฬ@1&2&3@3&3&1)| = ๐ ฬ (2(1) โ 3(3)) โ ๐ ฬ (1(1) โ 3(3)) + ๐ ฬ(1(3) โ 3(2)) = ๐ ฬ (2 โ 9) โ ๐ ฬ (1 โ 9) + ๐ ฬ(3 โ 6) = โ7๐ ฬ + 8๐ ฬ โ 3๐ ฬ Hence, direction ratios = โ7, 8, โ3 โด A = โ7, B = 8, C = โ3 Putting above values in (1) A(x + 1) + B (y โ 3) + C(z โ 2) = 0 โ7k (x + 1) + 8k (y โ 3) โ 3k (z โ2) = 0 k [โ7(๐ฅ+1)+8(๐ฆโ3)โ3(๐งโ2)] = 0 โ7(x + 1) + 8(y โ 3) โ 3(z โ 2) = 0 โ7x โ 7 + 8y โ 24 โ 3z + 6 = 0 โ7x + 8y โ3z โ 25 = 0 0 = 7x โ 8y + 3z + 25 7x โ 8y + 3z + 25 = 0 Therefore equation of the required plane is 7x โ 8y + 3z + 25 = 0 Question 9 (Method 2) Find the equation of the plane passing through the point (โ 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. The equation of a plane passing through (๐ฅ_1, ๐ฆ_1, ๐ง_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 (โ1, 3, 2) So, equation of plane is A(x + 1) + B (y โ 3) + 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 + 1) + B (y โ 3) + C(z โ 2) = 0 is perpendicular to plane x + 2y + 3z = 5 Hence, A ร 1 + B ร 2 + C ร 3 = 0 A + 2B + 3C = 0 Similarly, Given that plane A(x + 1) + B (y โ 3) + C(z โ 2) = 0 is perpendicular to plane 3x + 3y + z = 0. 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 3 + B ร 3 + C ร 1 = 0 3A + 3B + C = 0 So, our equations are A + 2B + 3C = 0 3A + 3B + C = 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 ๐ด/(2 โ 9) = ๐ต/(9 โ 1) = ๐ถ/(3 โ 6) ๐จ/(โ๐) = ๐ฉ/๐ = ๐ช/(โ๐) = k So, A = โ7k, B = 8k, C = โ3k Putting above values in (1) A(x + 1) + B (y โ 3) + C(z โ 2) = 0 โ7k (x + 1) + 8k (y โ 3) โ 3k (z โ2) = 0 k [โ7(๐ฅ+1)+8(๐ฆโ3)โ3(๐งโ2)] = 0 โ7(x + 1) + 8(y โ 3) โ 3(z โ 2) = 0 โ7x โ 7 + 8y โ 24 โ 3z + 6 = 0 โ7x + 8y โ3z โ 25 = 0 0 = 7x โ 8y + 3z + 25 7x โ 8y + 3z + 25 = 0 Therefore equation of the required plane is 7x โ 8y + 3z + 25 = 0