Find the image of the point(0,2,3) in the line (x+3)/5=(y-1)/2=(z+4)/3 . The answer in the book is: (4,4,-5).

Asked by Manoj | 13th May, 2013, 08:21: PM

Expert Answer:

Answer : Let P (0, 2, 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 1y 1z 1) be the image of point P. Then we will find the values of x 1y 1z 1.

 

Then,

Given line is

 

=>  (x+3) / 5 =  (y-1)/2  =  (z+4)/3  =t (let)

=> x = 5t -3 , y = 2t+1 , z= 3t - 4

i.e., the co-ordinate of any general point which lies on the given line is ( 5t -3 , 2t+1 , 3t - 4).

 

? Co-ordinate of L are ( 5t -3 , 2t+1 , 3t - 4). 

 

Now,

Direction ratio's of PL will be

 5t -3 -0 , 2t+1-2 , 3t - 4-3

= 5t - 3 , 2t -1 , 3t-7

  

Also,

Direction ratio's of given line are 5, 2, 3.

 

But, PL is perpendicular to the given line.

=> 5(5t - 3) +2( 2t -1 ) +3 ( 3t-7) =0

=> 38t = 38 

=> t = 1

 

? Co-ordinate of L are (2, 3, -1)

 

Now,

L is the mid point of PP',

By mid point formula

=> (0 + x1 )/2 = 2 , (2+ y1)/2 =3 , (3+ z1)/2 = -1

 =>  x1 = 4 ,  y1 =4 ,  z1 = -5 

Hence, the image of (0, 2, 3) is (4, 4, -5).

Answered by  | 13th May, 2013, 08:45: PM

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