Examples
Example, 2 Important
Example, 3
Example, 4 Important
Example, 5 Important
Example, 6 Important
Example, 7
Example 8 Important
Example 9
Example 10 Important
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 16 Deleted for CBSE Board 2024 Exams
Question 17 Important Deleted for CBSE Board 2024 Exams
Question 18 Important Deleted for CBSE Board 2024 Exams You are here
Question 19 Important Deleted for CBSE Board 2024 Exams
Question 20 Important Deleted for CBSE Board 2024 Exams
Last updated at April 16, 2024 by Teachoo
Question 18 Find the distance between the point P(6, 5, 9) and the plane determined by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6).The equation of a plane passing through points A(𝑥_1, 𝑦_1, 𝑧_1) B(𝑥_2, 𝑦_2, 𝑧_2) and C (𝑥_3, 𝑦_3, 𝑧_3) is |■8(𝒙−𝒙_𝟏&𝒚−𝒚_𝟏&𝒛−𝒛_𝟏@𝒙_𝟐−𝒙_𝟏&𝒚_𝟐−𝒚_𝟏&𝒛_𝟐−𝒛_𝟏@𝒙_𝟑−𝒙_𝟏&𝒚_𝟑−𝒚_𝟏&𝒛_𝟑−𝒛_𝟏 )| = 0 Given, the three points are A(3, −1, 2) 𝑥_1= 3, 𝑦_1= −1, 𝑧_1= 2 B(5, 2, 4) 𝑥_2= 5, 𝑦_2 = 2, 𝑧_2= 4 C(–1, –1, 6) 𝑥_3= −1, 𝑦_3= −1, 𝑧_3= 6 Equation of plane is |■8(𝑥−3&𝑦−(−1)&𝑧−2@5−3&2−(−1)&4−2@−1−3&−1−(−1)&6−2)| = 0 |■8(𝑥−3&𝑦+1&𝑧−2@2&3&2@−4&0&4)| = 0 (x − 3)[(3×4)−(0×2)] − (y + 1) [(2×4)−(−4×2)] + (z − 2) [(2×0)−(−4×3)] (x − 3)[12−0] − (y + 1) [8+8] +(𝑧−2) [0+12] = 0 12(x − 3) – 16(y + 1) + 12(z − 2) = 0 3(x − 3) – 4(y + 1) + 3(z − 2) = 0 3x − 9 – 4y – 4 + 3z − 6 = 0 3x – 4y + 3z − 19 = 0 Therefore, equation of plane is 3x – 4y + 3z = 19 Now, the distance between a point P(𝑥_1, 𝑦_1, 𝑧_1) and the plane Ax + By + Cz = D is |(𝑨𝒙_𝟏 + 𝑩𝒚_𝟏 + 𝑪𝒛_𝟏− 𝑫)/√(𝑨^𝟐 + 𝑩^𝟐 + 𝑪^𝟐 )| Given, the point is P(6, 5, 9) So, 𝑥_1= 6 , 𝑦_1= 5, 𝑧_1= 9 The equation of plane is 3x – 4y + 3z = 19 Comparing with Ax + By + Cz = D, A = 3, B = –4, C = 3, D = 19 Now, Distance of the point from the plane = |((3 × 6) + (−4 × 5) + (3 × 9) − 19)/√(3^2 + 4^2 + 3^2 )| =|(18 − 20 + 27 − 19)/√(9 + 16 + 9)| =|6/√34| = 6/√34 = 6/√34 × √34/√34 = (6√34)/34 = (𝟑√𝟑𝟒)/𝟏𝟕