HSC Arts 12th Board ExamMaharashtra State Board
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Find the Length of the Perpendicular from the Point (3, 2, 1) to the Line (x-7)/2=(y-7)/2=(z-6)/3=λ(say) - HSC Arts 12th Board Exam - Mathematics and Statistics

ConceptThree - Dimensional Geometry Condition for Perpendicular Lines

Question

Find the length of the perpendicular from the point (3, 2, 1) to the line (x-7)/2=(y-7)/2=(z-6)/3=lambda (say)

Solution

Let M be the foot of the perpendicular drawn from point A (3, 2, 1) to the given line

(x-7)/2=(y-7)/2=(z-6)/3=lambda (say)

therefore x=7-2λ,y=7+2λ,z=6+3λ

are the coordinates of any point on the given line

M=(7-2λ,y=7+2λ,6+3λ)

The d.r.s of AM are 7 - 2λ - 3, 7 + 2λ - 2, 6 + 3λ-1
i.e. 4 - 2λ, 5 + 2λ, 5 + 3λ
The d.r.s of the given line are -2, 2, 3
Since, AM is perpendicular to the given line
(4 - 2λ) (-2) + (5 + 2λ) (2) + (5 + 3λ) (3) = 0
-8 + 4λ + 10 + 4λ + 15 + 9λ = 0
17λ+ 17 = 0
λ = -1 [1]
The coordinates of the foot of the perpendicular

i.e. M = (9, 5, 3)
The length of the perpendicular distance

i.e. AM =  sqrt((9-3)^2+(5-2)^2+(3-1)^2)

= sqrt(49)

l(AM) = 7 units

Is there an error in this question or solution?

APPEARS IN

2014-2015 (October) (with solutions)
Question 3.1.3 | 3.00 marks
Solution Find the Length of the Perpendicular from the Point (3, 2, 1) to the Line (x-7)/2=(y-7)/2=(z-6)/3=λ(say) Concept: Three - Dimensional Geometry - Condition for Perpendicular Lines.
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