Distance of point from plane
Question 14 (a) Important Deleted for CBSE Board 2025 Exams
Question 18 Important Deleted for CBSE Board 2025 Exams
Question 17 (MCQ) Important Deleted for CBSE Board 2025 Exams
Question 16 Important Deleted for CBSE Board 2025 Exams
Question 10 Important Deleted for CBSE Board 2025 Exams
Distance of point from plane
Last updated at April 16, 2024 by Teachoo
Question 14 Find the distance of a point (2, 5, –3) from the plane 𝑟 . (6 𝑖 – 3 𝑗 + 2 𝑘) = 4 The distance of a point with position vector 𝑎 from the plane 𝑟. 𝑛 = d, where 𝑛 is the normal to the plane is 𝒂. 𝒏 − 𝒅 𝒏 Given, the point is (2, 5, −3) So, 𝑎 = 2 𝑖 + 5 𝑗 − 3 𝑘 The equation of plane is 𝑟.(6 𝑖 − 3 𝑗 + 2 𝑘) = 4 Comparing with 𝑟. 𝑛 = d, 𝑛 = 6 𝑖 − 3 𝑗 + 2 𝑘 & d = 4 Distance of point from plane = 𝑎. 𝑛 − 𝑑 𝑛 = 2 𝑖 + 5 𝑗 − 3 𝑘. 6 𝑖 − 3 𝑗 + 2 𝑘 − 4 62 + −32 + 22 = 2 × 6 + 5 × −3 + −3 × 2 − 4 36 + 9 + 4 = 12 − 15 − 6 − 4 49 = −137 = 𝟏𝟑𝟕