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 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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