### Find the distance of point − 2i + 3j − 4k from the line

### r = i + 2j − k + λ (i + 3j − 9k )

### measured parallel to the plane: x – y + 2z – 3 = 0.

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Last updated at Dec. 21, 2017 by Teachoo

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Question 28 Find the distance of point β2πΒ Μ + 3πΒ Μ β 4πΒ Μ from the line πΒ Μ = πΒ Μ + 2πΒ Μ β πΒ Μ + π (πΒ Μ + 3πΒ Μ β 9πΒ Μ) measured parallel to the plane: x β y + 2z β 3 = 0. Let Point A = β2πΒ Μ + 3πΒ Μ β 4πΒ 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 πΒ Μ = πΒ Μ + 2πΒ Μ β πΒ Μ + π (πΒ Μ + 3πΒ Μ β 9πΒ 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 πΒ Μ β πΒ Μ + 2πΒ 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 πΒ Μ β πΒ Μ + 2πΒ Μ Direction cosines of AB = (π+3, 3πβ 1, β9π+3) Direction cosines of πΒ Μ β πΒ Μ + 2πΒ 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

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