Question

Find the coordinates of the foot of perpendicular drawn from the point (-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.

Answer 1

hope you got the answer

Answer 2

Given:

Point A (-1, 8, 4).

Line joining the points B (0, -1, 3) and C (2, -3, -1)

To Find:

The image of the point A in the line BC.

Solution:

Let P( x , y , z) be the foot of the perpendicular.

• Parallel vector to BC = 2i - 2j -4k

• Equation of line BC =  -j +3k  + t(2i - 2j -4k)

Any point on line BC  =( 2t , -1 - 2t , 3 - 4t )

Parallel vector to AP = (x + 1) i + (y -8) j + (z - 4) k

Since they are perpendicular ,

• BC.AP = 0
• <2, -2, -4>.<x+1 , y-8, z-4> = 0
• 2 (2t) + 2 -2( -1 -2t) +16 - 4( 3 - 4t) + 16 = 0
• 4t + 2 + 2 +4t + 16 -12 + 16t + 16 = 0
• 24t + 24 = 0
• t = -1

Foot of the perpendicular ,

• P = ( -2 , 1 , 7 )
• Parallel vector to AP = -i -7j + 3k

Let M ( p, q, r ) be the mirror of A .

• M lies on the line joining P and A.

Since M is the mirror image of A , P is the mid point of line AM

• (-2,1,7) = (M + A) /2 = (p  -1 , q + 8 , r + 4) /2
• ( -4, 2 , 14 ) = ( p - 1, q + 8, r + 4 )
• p = -3 , q = -6 , r = 10
• point M is ( -3, -6, 10)

The coordinates of the foot of perpendicular drawn from the point (-1, 8, 4) to the line joining the points B (0, -1, 3) and C (2, -3, -1) is ( -2, 1, 7 ).

The image of the point A in the line BC in ( -3, -6 ,10 ).