Question 37 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - 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 drawn from the point (-1, 3, -6) to the plane 2π₯ + π¦ − 2π§ + 5 = 0. Also find the equation and length of the perpendicular.

Note
: This
is similar
to
Example 16
of NCERT –
Chapter 11 Class 12 Three Dimensional Geometry

Question 37 (Choice 2) Find the foot of the perpendicular drawn from the point (β1, 3, β6) to the plane 2π₯ + π¦ β 2π§ + 5 = 0. Also find the equation and length of the perpendicular.
Let point P(x1, y1, z1) be foot of perpendicular from point X (β1, 3, β6)
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
2x + y β 2z + 5 = 0
2x + y β 2z = β5
So, Normal vector = π β = 2π Μ + π Μ β 2π Μ
Since, (πΏπ·) β and π β are parallel
their direction ratios are proportional.
Finding direction ratios
(πΏπ·) β = (x1 + 1)π Μ + (y1 β 3)π Μ + (z1 + 6)π Μ
Direction ratios = x1 + 1, y1 β 3, z1 + 6
β΄ a1 = x1 + 1 , b1 = y1 β 3, c1 = z1 + 6
π β = 2π Μ + π Μ β 2π Μ
Direction ratios = 2, 1, β2
β΄ a2 = 2 , b2 = 1, c2 = β2
Direction ratios are proportional
π_1/π_2 = π_1/π_2 = π_1/π_2 = k
(π₯_1 + 1)/2 = (π¦_1 β 3)/( 1) = (π§_1 + 6)/(β2) = k
Thus,
x1 = 2k β 1,
y1 = k + 3,
z1 = β2k β 6
Also, point P(x1, y1, z1) lies in the plane.
Putting P (2k β 1, k + 3, β2k β 6) in equation of plane
2x + y β 2z = β5
2(2k β 1) + (k + 3) β 2(β2k β 6) = β5
4k β 2 + k + 3 + 4k + 12 = β5
4k + k + 4k β 2 + 3 + 12 = β5
9k + 13 = β5
9k = β5 β 13
9k = β18
β΄ k = β2
Thus,
x1 = 2k β 1 = 2(β2) β 1 = β5
y1 = k + 3 = (β2) + 3 = 1
z1 = β2k β 6 = β2(β2) β 6 = β2
Therefore, coordinate of foot of perpendicular are P (β5, 1, β2)
Equation of perpendicular
Equation of perpendicular would be equation of line joining X (β1, 3, β6) and P (β5, 1, β2)
(π₯ β (β1))/(β5 β (β1))=(π¦ β 3)/(1 β 3)=(π§ β (β6))/(β2 β (β6))
(π₯ + 1)/(β4)=(π¦ β 3)/(β2)=(π§ + 6)/4
(π + π)/(βπ)=(π β π)/(βπ)=(π + π)/π
Length of perpendicular
X (β1, 3, β6) and P (β5, 1, β2)
Let of Perpendicular is length of PX
PX = β((β5β(β1))^2+(1β3)^2+(β2β(β6))^2 )
PX = β((β5+1)^2+(β2)^2+(β2+6)^2 )
PX = β((β4)^2+(β2)^2+(4)^2 )
PX = β(16+4+16)
PX = β36
PX = 6 units

Made by

Davneet Singh

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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