Question 28
Find the distance of point β2π Μ + 3π Μ β 4π Μ from the line
π Μ = π Μ + 2π Μ β π Μ + π (π Μ + 3π Μ β 9π Μ)
measured parallel to the plane: x β y + 2z β 3 = 0.
Let Point A = β2π Μ + 3π Μ β 4π Μ
β΄ A = (β2, 3, β4)
Let point B the point on line such that AB is parallel to the plane
Equation of line is
π Μ = π Μ + 2π Μ β π Μ + π (π Μ + 3π Μ β 9π Μ)
In Cartesian form,
(π₯ β 1)/1=(π¦ β 2)/3=(π§ + 1)/(β9)
Since point B lies on line
Itβs coordinates will be
(π₯ β 1)/1=(π¦ β 2)/3=(π§ + 1)/(β9)=π
So,
x = π+1
y = 3π+2
z = β9πβ1
So, B = (π+1, 3π+2, β9πβ1)
Now,
Line AB is parallel to the plane
So, it will be perpendicular to planeβs normal
β΄ Line AB is perpendicular to π Μ β π Μ + 2π Μ
Finding equation of line AB
Equation of a line passing through two points (x1, y1, z1) and (x1, y1, z1)
(π₯ β π₯1)/(π₯2 β π₯1) = (π¦ β π¦1)/(π¦2 β π¦1) = (π§ β π§1)/(π§2 β π§1)
We use (x1, y1, z1) = Point A = (β2, 3, β4)
& (x2, y2, z2) = B = (π+1, 3π+2, β9πβ1)
So, equation of AB is
(π₯ β (β2))/((π+1) β (β2)) = (π¦ β 3)/((3π+2) β (3)) = (π§ β (β4))/((β9πβ1) β (β4))
(π₯ + 4)/(π + 3) = (π¦ β 3)/(3π β 1) = (π§ + 4)/(β9π + 3)
From (1)
Line AB is perpendicular to π Μ β π Μ + 2π Μ
Direction cosines of AB = (π+3, 3πβ 1, β9π+3)
Direction cosines of π Μ β π Μ + 2π Μ = (1, β1, 2)
Thus,
(π+ 3) Γ 1 + (3πβ 1) Γ β1 + (β9π+3) Γ 2 = 0
π+ 3 β 3π + 1 β18π+ 6 = 0
3 + 1 + 6 = 18π+ 3π β π
10 = 20π
10/20 = π
π = 1/2
So, point B is
B = (π+1, 3π+2, β9πβ1)
= (1/2 + 1, 3(1/2)+2, β9(1/2)β1)
= (3/2, 7/2,(β11)/2)
Thus,
Distance between A(β2, 3, β4) and B (3/2, 7/2,(β11)/2)
AB = β((3/2β(β2))^2+(7/2β3)^2+((β11)/2β(β4))^2 )
= β((3/2+2)^2+(7/2β3)^2+((β11)/2+4)^2 )
= βππ/π unit

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.