Find coordinates of the point on line x-1/2=y+2/3=z-3/6 which are distance 3 unit from the point (1,-2,3)

Asked by Sandeepkdaksh | 2nd Jan, 2019, 08:28: AM

Expert Answer:

Given line begin mathsize 12px style fraction numerator x minus 1 over denominator 2 end fraction space equals space fraction numerator y plus 2 over denominator 3 end fraction space equals space fraction numerator z minus 3 over denominator 6 end fraction end style
hence point on the line satisfy the condition, begin mathsize 12px style fraction numerator x minus 1 over denominator 2 end fraction space equals space fraction numerator y plus 2 over denominator 3 end fraction space equals space fraction numerator z minus 3 over denominator 6 end fraction space equals space k space space space o r space space space space space left parenthesis x comma y comma z right parenthesis space equals space left parenthesis 2 k plus 1 comma space 3 k minus 2 comma space 6 k plus 3 right parenthesis end style..............(1)
where k is a constant
If distance between a point on the line and a point (1,-2,3) is 3 unit , then we have
 
begin mathsize 12px style square root of open parentheses 2 k plus 1 minus 1 close parentheses squared plus open parentheses 3 k minus 2 plus 2 close parentheses squared plus left parenthesis 6 k plus 3 minus 3 right parenthesis squared space end root equals space space 3 end style   ..................................(2)
solving the above eqn.(2) for k, we get k = 3/7
 
hence from (1), we get the coordinates of required point as (13/7 , -5/7, 39/7)

Answered by Thiyagarajan K | 2nd Jan, 2019, 11:31: AM

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