Q. 27

Mark against the A. (3, 1, 8)

B. (1, 2, 8)

C. (3, -3, 5)

D. (5, -3, -4)

Answer :

Given: Perpendicular dropped from A(7, 14, 5) on to the plane 2x + 4y – z = 2


To find: co-ordinates of the foot of perpendicular


Formula Used: Equation of a line is



Where b1:b2:b3 is the direction ratio and (x1, x2, x3) is a point on the line.


Explanation:


Let the foot of the perpendicular be (a, b, c)


Since this point lies on the plane,


2a + 4b – c = 2 … (1)


Direction ratio of the normal to the plane is 2 : 4 : -1


Direction ratio perpendicular = direction ratio of normal to the plane


So, equation of the perpendicular is



Since (a, b, c) is a point on the perpendicular,



(7, 14, 5) is a point on the perpendicular.



So, a = 7 - 2λ, b = 14 – 4λ, c = 5 + λ


Substituting in (1),


14 – 4λ + 56 – 16λ – 5 - λ = 2


21λ = 70 – 7 = 63


λ = 3


Therefore, foot of the perpendicular is (1, 2, 8)

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