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 April 16, 2024 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 β 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 (πΏπ·) β is parallel to normal vector π β
Given equation of the plane is
x β 3y + 2z = 9
So, Normal vector = π β = π Μ β 3π Μ + 2π Μ
Since, (πΏπ·) β and π β are parallel
their direction ratios are proportional.
Finding direction ratios
(πΏπ·) β = (x1 β 1)π Μ + (y1 β 2)π Μ + (z1 β 0)π Μ
Direction ratios = x1 β 1, y1 β 2, z1
β΄ a1 = x1 β 1 , b1 = y1 β 2, c1 = z1
π β = 1π Μ β 3π Μ + 2π Μ
Direction ratios = 1, β3, 2
β΄ a2 = 1 , b2 = β3, c2 = 2
Since, (πΏπ·) β and π β are parallel
their direction ratios are proportional.
Finding direction ratios
Direction ratios are proportional
π_1/π_2 = π_1/π_2 = π_1/π_2 = k
(π₯_1 β 1)/1 = (π¦_1 β 2)/( β3) = π§_1/2 = k
Thus,
x1 = k + 1,
y1 = β3k + 2,
z1 = 2k
Also, point P(x1, y1, z1) lies in the plane.
Putting P (k + 1, β3k + 2, 2k) in equation of plane
x β 3y + 2z = 9
(k + 1) β 3(β3k + 2) + 2(2k) = 9
k + 1 + 9k β 6 + 4k = 9
k + 9k + 4k + 1 β 6 = 9
14k β 5 = 9
14k = 9 + 5
14k = 14
β΄ k = 1
Thus,
x1 = k + 1 = 1 + 1 = 2
y1 = β3k + 2 = β3(1) + 2 = β1
z1 = 2k = β2(1) = 2
Therefore, coordinate of foot of perpendicular are P (2, β1, 2)
Length of perpendicular
X (1, 2, 0) and P (2, β1, 2)
Let of Perpendicular is length of PX
PX = β((2β1)^2+(β1β2)^2+(2β0)^2 )
PX = β(1^2+(β3)^2+2^2 )
PX = β(1+9+4)
PX = βππ units

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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