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 Oct. 27, 2020 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

Check the answer here

https://www.teachoo.com/3572/756/Example-16---Find-coordinates-of-foot-of-perpendicular-from/category/Examples/

Transcript

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

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