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).
Asked by Manoj
| 13th May, 2013,
08:28: PM
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).
Answered by
| 13th May, 2013,
09:00: PM
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