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

Last updated at March 16, 2023 by Teachoo

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.

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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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