Examples

Chapter 11 Class 12 Three Dimensional Geometry
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Example 27 (Method 1) Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. The equation of a plane passing through (𝑥_1, 𝑦_1, 𝑧_1) is given by A(x − 𝒙_𝟏) + B (y − 𝒚_𝟏) + C(z – 𝒛_𝟏) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (1, −1, 2) So, equation of plane is A(x −1) + B (y + 1) + C(z − 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 𝑎 ⃗ × 𝑏 ⃗ is perpendicular to both 𝑎 ⃗ & 𝑏 ⃗ So, required is normal is cross product of normal of planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Required normal = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@2&3&−2@1&2&−3)| = 𝑖 ̂ (3(–3) – 2(–2)) – 𝑗 ̂ (2(–3) – 1(–2)) + 𝑘 ̂(2(2) – 1(3)) = 𝑖 ̂ (–9 + 4) – 𝑗 ̂ (–6 + 2) + 𝑘 ̂(4 – 3) = –5𝑖 ̂ + 4𝑗 ̂ + 𝑘 ̂ Hence, direction ratios = –5, 4, 1 ∴ A = –5, B = 4, C = 1 Putting above values in (1), A(x −1) + B (y + 1) + C(z − 2) = 0 −5(x − 1) + 4 (y + 1) + 1 (z − 2) = 0 −5x + 5 + 4y + 4 + z − 2 = 0 −5x + 4y + z + 7 = 0 −5x + 4y + z = −7 −(5x −4y − z) = −7 5x − 4y − z = 7 Therefore, the equation of the required plane is 5x − 4y − z = 7. Example 27 (Method 2) Find the equation of the plane that contains the point (1, – 1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. The equation of a plane passing through (𝑥_1, 𝑦_1, 𝑧_1) is given by A(x − 𝒙_𝟏) + B (y − 𝒚_𝟏) + C(z – 𝒛_𝟏) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (1, −1, 2) So, equation of plane is A(x −1) + B (y + 1) + C(z − 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 + 1) + C(z − 2) = 0 is perpendicular to plane 2x + 3y – 2z = 5 Hence, A × 2 + B × 3 + C × (–2) = 0 2A + 3B − 2C = 0 Similarly, Given that plane A(x −1) + B (y + 1) + C(z − 2) = 0 is perpendicular to plane x + 2y – 3z = 8 Two lines with direction ratios 𝑎_1, 𝑏_1, 𝑐_1 and 𝑎_2, 𝑏_2, 𝑐_2 are perpendicular if 𝑎_1 𝑎_2 + 𝑏_1 𝑏_2 + 𝑐_1 𝑐_2 = 0 Hence, A × 1 + B × 2 + C × (–3) = 0 A + 2B − 3C = 0 So, our equations are 2A + 3B −2C = 0 A + 2B − 3C = 0 Solving Two lines with direction ratios 𝑎_1, 𝑏_1, 𝑐_1 and 𝑎_2, 𝑏_2, 𝑐_2 are perpendicular if 𝑎_1 𝑎_2 + 𝑏_1 𝑏_2 + 𝑐_1 𝑐_2 = 0 𝐴/(−9 − (−4)) = 𝐵/(−2 − (−6)) = 𝐶/(4 − 3) 𝐴/(−9 + 4) = 𝐵/(−2 + 6) = 𝐶/1 𝐴/(−5) = 𝐵/4 = 𝐶/1 = k So, A = –5k , B = 4k , C = k Putting above values in (1), A(x −1) + B (y + 1) + C(z − 2) = 0 −5k(x − 1) + 4k (y + 1) + k (z − 2) = 0 k[−5(x − 1) + 4(y + 1) + (z − 2)] = 0 −5x + 5 + 4y + 4 + z − 2 = 0 −5x + 4y + z + 7 = 0 −5x + 4y + z = −7 −(5x −4y − z) = −7 5x − 4y − z = 7 Therefore, the equation of the required plane is 5x − 4y − z = 7. 