Find the image of the point(5,9,3) in the line (x-1)/2=(y-2)/3=(z-3)/4 The answer in the book is: (1,1,11).

Answer : Let P (5, 9, 3) be the point whose image is to be determined.

 

Draw perpendicular from P to the given line which cuts the line L, which is its foot of perpendicular.

 

Then, image of P will be in the downward direction and at the same distance from L as P.

 

Let P' (x ₁, y ₁, z ₁) be the image of point P. Then we will find the values of x ₁, y ₁, z ₁.

 

Then,

Given line is

 

=> (x-1)/2=(y-2)/3=(z-3)/4 =t (let)

=> x = 2t +1, y = 3t+2 , z= 4t +3

i.e., the co-ordinate of any general point which lies on the given line is ( 2t +1, 3t+2 , 4t +3).

 

? Co-ordinate of L are ( 2t +1, 3t+2 , 4t +3). 

 

Now,

Direction ratio's of PL will be

   2t +1-5 , 3t+2 -9 , 4t +3 -3

= 2t-4 , 3t-7 , 4t 

  

Also,

Direction ratio's of given line are 2,3, 4.

 

But, PL is perpendicular to the given line.

=> 2 (2t-4) + 3( 3t-7 )+ 4( 4t ) =0  

=> 29 t = 29  

=> t = 1 

 

? Co-ordinate of L are (3, 5, 7)

 

Now,

L is the mid point of PP',

By mid point formula

=> (5+ x₁ )/2 = 3 , (9+ y₁)/2 =5 , (3+ z₁)/2 = 7

 =>  x₁ = 1 ,  y₁ = 1 ,  z₁ = 11 

 

Hence, the image of (5, 9, 3) is (1, 1, 11).