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  • 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.   &nb

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.

 

 

 

 
 
 
 
 

Answers (1)

Equation of line :

BC: \frac{x-0}{2-0}=\frac{y-(-1)}{-3-(-1)}=\frac{z-3}{-1-3}

i.e.\; \; \frac{x}{2}=\frac{y+1}{-2}=\frac{z-3}{-4}=\lambda  (say)

\therefore  Coordinates of random point on this line is P(2\lambda ,-2\lambda -1,-4\lambda +3)

The directions od AD are 2\lambda+1 ,-2\lambda -9,-4\lambda -1

As  AP\perp BC, so 

2(2\lambda +1)-2(-2\lambda -9)-4(-4\lambda -1)=0

\Rightarrow \lambda =-1

\therefore P(-2,1,7)

Let Q(h,p,s) be the image of A in the line BC. So P must be the mid point of AQ. 

\therefore P\left ( \frac{h-1}{2},\frac{p+8}{2},\frac{s+4}{2} \right )=P(-2,1,7)

Comparing the coordinates, we get 

h=-3,p=-6,s=10

Hence the image is Q(-3,-6,10)

 

 

Posted by

Ravindra Pindel

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