

Last updated at May 29, 2018 by Teachoo
Transcript
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 ( 𝑥1, 𝑦1, 𝑧1) is given by A(x − 𝒙𝟏) + B (y − 𝒚𝟏) + C(z – 𝒛𝟏) = 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 normals of both planes. We know that 𝑎 × 𝑏 is perpendicular to both 𝑎 & 𝑏 So, required is normal is cross product of normals of planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Required normal = 𝑖 𝑗 𝑘23−212−3 = 𝑖 (3(–3) – 2(–2)) – 𝑗 (2(–3) – 1(–2)) + 𝑘(2(2) – 1(3)) = 𝑖 (–9 + 4) – 𝑗 (–6 + 2) + 𝑘(4 – 3) = –5 𝑖 + 4 𝑗 + 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 ( 𝑥1, 𝑦1, 𝑧1) is given by A(x − 𝒙𝟏) + B (y − 𝒚𝟏) + C(z – 𝒛𝟏) = 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 Hence, A × 1 + B × 2 + C × (–3) = 0 A + 2B − 3C = 0 So, our equations are 2A + 3B −2C = 0 …(2) A + 2B − 3C = 0 …(3) Solving 𝐴−9 − (−4) = 𝐵−2 − (−6) = 𝐶4 − 3 𝐴−9 + 4 = 𝐵−2 + 6 = 𝐶1 𝐴−5 = 𝐵4 = 𝐶1 = 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.
Examples
Example, 2
Example, 3 Important
Example, 4
Example, 5
Example, 6 Important
Example, 7
Example 8
Example, 9 Important
Example 10
Example 11
Example 12 Important
Example 13
Example 14
Example 15
Example 16
Example 17
Example 18
Example 19
Example 20 Important
Example 21 Important
Example 22
Example 23 Important
Example 24 Important
Example, 25 Important
Example 26
Example 27 Important You are here
Example 28
Example 29 Important
Example 30 Important
About the Author