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CBSE Class 12-science Answered

Can you send me the solution of 39
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Asked by divakar_sabu | 15 Dec, 2018, 01:27: PM
answered-by-expert Expert Answer
Given Line is begin mathsize 12px style fraction numerator x plus 2 over denominator 3 end fraction space equals space fraction numerator y plus 1 over denominator 2 end fraction space equals space fraction numerator z minus 3 over denominator 2 end fraction space equals space k
end style.   
Then points on the line is in the form (3k-2, 2k-1, 2k+3)......................(1)

distance between the line and the point (1,2,3) = 3√2 . Hence we have,
 
begin mathsize 12px style square root of open parentheses 3 k minus 2 minus 1 close parentheses squared plus open parentheses 2 k minus 1 minus 2 close parentheses squared plus open parentheses 2 k plus 3 minus 3 close parentheses squared space end root space equals space 3 square root of 2
9 open parentheses k minus 1 close parentheses squared plus space open parentheses 2 k minus 3 close parentheses squared plus 4 k squared space equals space 18
17 k squared minus 30 k plus 18 space equals space 18 space space
k left parenthesis 17 k minus 30 right parenthesis space equals space 0 space space o r space space k space equals space 0 space space o r space k space equals space 30 over 17 end style
 
Hence points on the line are obtained by substituting k in eqn.(1).

The required points are (-2, -1, 3) if k=0  or [ (56/17), (43/17), (111/17) ] if k = 30/17
Answered by Thiyagarajan K | 17 Dec, 2018, 06:15: AM
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