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Question
Find the coordinates of the foot of perpendicular drawn from the point A (-1,8,4) to the line joining the points B(0,-1,3) and C(2,-3,-1). Hence find the image of the point A in the line BC.
Solution
Let P be the foot of the perpendicular drawn from point A on the line joiningpoints B and C.
Let P’ (a, b, c) be the coordinates of image of point A.
Equation of line BC is given by,
`(x-x_1)/(x_2-x_1)=(y-y_1)/(y_2-y_1)=(z-z_1)/(z_2-z_1)`
`(x-0)/2=(y+1)/(-2)=(z-3)/(-4)=lambda`
General coordinates of P is (2λ,-2λ-1,-4λ+3)
Direction ratios of AP (2λ+1,-2λ-9,-4λ-1)
AP⊥BC
2(2λ+1)-2(2λ-9)-4(-4λ-1)=0
4λ+2+4λ+18+16λ+4=0
24+24λ=0
λ=-1
P(-2,1,7)
Coordinates of foot of perpendicular is (-2,1,7)
Coordinates of image of A is P' (a, b, c) is
`(a-1)/2=-2, a=-3`
`(b+8)/2=1, b= -6`
`(c+4)/2=7, c=10`
P'(-3,-6,10)
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