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Examples

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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at June 3, 2021 by Teachoo

Example 27 (Method 1) Find the equation of the plane that contains the point (1, β1, 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, β1, 2) So, equation of plane is A(x β1) + 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&β2@1&2&β3)| = π Μ (3(β3) β 2(β2)) β π Μ (2(β3) β 1(β2)) + π Μ(2(2) β 1(3)) = π Μ (β9 + 4) β π Μ (β6 + 2) + π Μ(4 β 3) = β5π Μ + 4π Μ + π Μ Hence, direction ratios = β5, 4, 1 β΄ A = β5, B = 4, C = 1 Putting above values in (1), A(x β1) + B (y + 1) + C(z β 2) = 0 β5(x β 1) + 4 (y + 1) + 1 (z β 2) = 0 β5x + 5 + 4y + 4 + z β 2 = 0 β5x + 4y + z + 7 = 0 β5x + 4y + z = β7 β(5x β4y β z) = β7 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, β1, 2) So, equation of plane is A(x β1) + 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 β1) + 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 (β2) = 0 2A + 3B β 2C = 0 Similarly, Given that plane A(x β1) + 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 (β3) = 0 A + 2B β 3C = 0 So, our equations are 2A + 3B β2C = 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΄/(β9 β (β4)) = π΅/(β2 β (β6)) = πΆ/(4 β 3) π΄/(β9 + 4) = π΅/(β2 + 6) = πΆ/1 π΄/(β5) = π΅/4 = πΆ/1 = k So, A = β5k , B = 4k , C = k Putting above values in (1), A(x β1) + B (y + 1) + C(z β 2) = 0 β5k(x β 1) + 4k (y + 1) + k (z β 2) = 0 k[β5(x β 1) + 4(y + 1) + (z β 2)] = 0 β5x + 5 + 4y + 4 + z β 2 = 0 β5x + 4y + z + 7 = 0 β5x + 4y + z = β7 β(5x β4y β z) = β7 5x β 4y β z = 7 Therefore, the equation of the required plane is 5x β 4y β z = 7.