Find the foot of the perpendicular from the point (1, 2, 0) upon the plane x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given plane.

This question is similar to Question 37 (Choice 2) CBSE Class 12 - Sample Paper 2021 Boards

[Term 2] Find foot of perpendicular from point (1, 2, 0) upon plane - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)

part 2 - Question 13 - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 13 - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 4 - Question 13 - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 5 - Question 13 - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 13 Find the foot of the perpendicular from the point (1, 2, 0) upon the plane x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given plane. Let point P(x1, y1, z1) be foot of perpendicular from point X (1, 2, 0) Since perpendicular to plane is parallel to normal vector Vector (𝑿𝑷) ⃗ is parallel to normal vector 𝒏 ⃗ Given equation of the plane is x − 3y + 2z = 9 So, Normal vector = 𝒏 ⃗ = 𝒊 ̂ − 3𝒋 ̂ + 2𝒌 ̂ Since, (𝑿𝑷) ⃗ and 𝒏 ⃗ are parallel their direction ratios are proportional. Finding direction ratios (𝑿𝑷) ⃗ = (x1 − 1)𝒊 ̂ + (y1 − 2)𝒋 ̂ + (z1 − 0)𝒌 ̂ Direction ratios = x1 − 1, y1 − 2, z1 ∴ a1 = x1 − 1 , b1 = y1 − 2, c1 = z1 𝒏 ⃗ = 1𝒊 ̂ − 3𝒋 ̂ + 2𝒌 ̂ Direction ratios = 1, −3, 2 ∴ a2 = 1 , b2 = −3, c2 = 2 Since, (𝑿𝑷) ⃗ and 𝒏 ⃗ are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional 𝑎_1/𝑎_2 = 𝑏_1/𝑏_2 = 𝑐_1/𝑐_2 = k (𝑥_1 − 1)/1 = (𝑦_1 − 2)/( −3) = 𝑧_1/2 = k Thus, x1 = k + 1, y1 = −3k + 2, z1 = 2k Also, point P(x1, y1, z1) lies in the plane. Putting P (k + 1, −3k + 2, 2k) in equation of plane x − 3y + 2z = 9 (k + 1) − 3(−3k + 2) + 2(2k) = 9 k + 1 + 9k − 6 + 4k = 9 k + 9k + 4k + 1 − 6 = 9 14k − 5 = 9 14k = 9 + 5 14k = 14 ∴ k = 1 Thus, x1 = k + 1 = 1 + 1 = 2 y1 = −3k + 2 = −3(1) + 2 = −1 z1 = 2k = −2(1) = 2 Therefore, coordinate of foot of perpendicular are P (2, −1, 2) Length of perpendicular X (1, 2, 0) and P (2, −1, 2) Let of Perpendicular is length of PX PX = √((2−1)^2+(−1−2)^2+(2−0)^2 ) PX = √(1^2+(−3)^2+2^2 ) PX = √(1+9+4) PX = √𝟏𝟒 units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo