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Question: The co-ordinates of the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the...

The co-ordinates of the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the points B(4, 7, 1) and C(3, 5, 3) are

A

(5/3, 7/3, 17/3)

B

(5, 7, 17)

C

(5/3, –7/3, 17/3)

D

(–5/3, 7/3, –17/3)

Answer

(5/3, 7/3, 17/3)

Explanation

Solution

Equation of BC, x−4−1=y−7−2=z−12\frac { x - 4 } { - 1 } = \frac { y - 7 } { - 2 } = \frac { z - 1 } { 2 }

i.e. x−41=y−72=z−1−2=r\frac { x - 4 } { 1 } = \frac { y - 7 } { 2 } = \frac { z - 1 } { - 2 } = r(say)

Any point on the given line is D(r+4,2r+7,−2r+1)D ( r + 4,2 r + 7 , - 2 r + 1 )

Then, d.r.’s of AD = (r+4−1,2r+7−0,−2r+1−3)( r + 4 - 1,2 r + 7 - 0 , - 2 r + 1 - 3 )

i.e. d.r.’s of AD=(r+3,2r+7,−2r−2)A D = ( r + 3,2 r + 7 , - 2 r - 2 ) and

d.r.’s of BC = (–1, –2, 2)

Since AD is ⊥ to given line,

∴ (−1)(r+3)+(2r+7)(−2)+(2)(−2r−2)=0( - 1 ) ( r + 3 ) + ( 2 r + 7 ) ( - 2 ) + ( 2 ) ( - 2 r - 2 ) = 0

⇒ −r−3−4r−14−4r−4=0- r - 3 - 4 r - 14 - 4 r - 4 = 0 ⇒ −9r−21=0- 9 r - 21 = 0

⇒ r=−7/3r = - 7 / 3

∴ D is {4 – (7/3), 7– (14/3), (14/3)+1}

i.e. D is (5/3, 7/3, 17/3)