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Perpendiculars are drawn from points on the line `(x+2)/2=(y+1)/(-1)=z/3` to the plane `x + y + z=3` The feet of perpendiculars lie on the line (a) `x/5=(y-1)/8=(z-2)/(-13)` (b) `x/2=(y-1)/3=(z-2)/(-5)` (c) `x/4=(y-1)/3=(z-2)/(-7)` (d) `x/2=(y-1)/(-7)=(z-2)/5`

A

`(x)/(5)= (y-1)/(8) = (z-2)/(-13)`

B

`(x)/(2)= (y+1)/(3)=(z-2)/(-5)`

C

`(x)/(4)= (y-1)/(3)= (z-2)/(-7)`

D

`(x)/(2)= (y-1)/(-7)= (z-2)/(5)`

Text Solution

Verified by Experts

Any point B on line is `(2lamda-2, -lamda-1, 3lamda)`
Point B lies on the plane for some `lamda`
`rArr" "(2lamda-2)+ (-lamda-1)+ 3lamda=3 or lamda = 3//2`
`rArr" "B-= (1, -5//2, 9//2)`
The foot of the perpendicular from the point `(-2, -1, 0)` on the plane is the point `A(0, 1, 2)`
`rArr" "` Direction ratio of `AB= (1, (-7)/(2), (5)/(2))-= (2, -7, 5)`.
Hence, feet of perpendicular lies on the line
`(x)/(2)= (y-1)/(-7)= (z-2)/(5)`
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Knowledge Check

  • The angle between the lines (x-2)/(3)=(y+1)/(-2),z=2and(x-1)/(1)=(2y=3)/(3)=(z+5)/(2)

    A
    `(pi)/(6)`
    B
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    C
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    D
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    A
    parallel
    B
    perpendicular
    C
    inclined at `45^(@)`
    D
    none

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