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 Jan. 25, 2022 by Teachoo

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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.

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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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.