Examples

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Example 27 (Method 1) Find the equation of the plane that contains the point (1, β1, 2) and is perpendicular to each of the planes 2x + 3y β 2z = 5 and x + 2y β 3z = 8. The equation of a plane passing through (π₯_1, π¦_1, π§_1) is given by A(x β π_π) + B (y β π_π) + C(z β π_π) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (1, β1, 2) So, equation of plane is A(x β1) + B (y + 1) + C(z β 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 β Γ π β is perpendicular to both π β & π β So, required is normal is cross product of normal of planes 2x + 3y β 2z = 5 and x + 2y β 3z = 8. Required normal = |β 8(π Μ&π Μ&π Μ@2&3&β[email protected]&2&β3)| = π Μ (3(β3) β 2(β2)) β π Μ (2(β3) β 1(β2)) + π Μ(2(2) β 1(3)) = π Μ (β9 + 4) β π Μ (β6 + 2) + π Μ(4 β 3) = β5π Μ + 4π Μ + π Μ Hence, direction ratios = β5, 4, 1 β΄ A = β5, B = 4, C = 1 Putting above values in (1), A(x β1) + B (y + 1) + C(z β 2) = 0 β5(x β 1) + 4 (y + 1) + 1 (z β 2) = 0 β5x + 5 + 4y + 4 + z β 2 = 0 β5x + 4y + z + 7 = 0 β5x + 4y + z = β7 β(5x β4y β z) = β7 5x β 4y β z = 7 Therefore, the equation of the required plane is 5x β 4y β z = 7. Example 27 (Method 2) Find the equation of the plane that contains the point (1, β 1, 2) and is perpendicular to each of the planes 2x + 3y β 2z = 5 and x + 2y β 3z = 8. The equation of a plane passing through (π₯_1, π¦_1, π§_1) is given by A(x β π_π) + B (y β π_π) + C(z β π_π) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (1, β1, 2) So, equation of plane is A(x β1) + B (y + 1) + C(z β 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 β1) + B (y + 1) + C(z β 2) = 0 is perpendicular to plane 2x + 3y β 2z = 5 Hence, A Γ 2 + B Γ 3 + C Γ (β2) = 0 2A + 3B β 2C = 0 Similarly, Given that plane A(x β1) + B (y + 1) + C(z β 2) = 0 is perpendicular to plane x + 2y β 3z = 8 Two lines with direction ratios π_1, π_1, π_1 and π_2, π_2, π_2 are perpendicular if π_1 π_2 + π_1 π_2 + π_1 π_2 = 0 Hence, A Γ 1 + B Γ 2 + C Γ (β3) = 0 A + 2B β 3C = 0 So, our equations are 2A + 3B β2C = 0 A + 2B β 3C = 0 Solving Two lines with direction ratios π_1, π_1, π_1 and π_2, π_2, π_2 are perpendicular if π_1 π_2 + π_1 π_2 + π_1 π_2 = 0 π΄/(β9 β (β4)) = π΅/(β2 β (β6)) = πΆ/(4 β 3) π΄/(β9 + 4) = π΅/(β2 + 6) = πΆ/1 π΄/(β5) = π΅/4 = πΆ/1 = k So, A = β5k , B = 4k , C = k Putting above values in (1), A(x β1) + B (y + 1) + C(z β 2) = 0 β5k(x β 1) + 4k (y + 1) + k (z β 2) = 0 k[β5(x β 1) + 4(y + 1) + (z β 2)] = 0 β5x + 5 + 4y + 4 + z β 2 = 0 β5x + 4y + z + 7 = 0 β5x + 4y + z = β7 β(5x β4y β z) = β7 5x β 4y β z = 7 Therefore, the equation of the required plane is 5x β 4y β z = 7.