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CBSE Class 12-science Answered

Find the shortest distance whose equations are r = (i-j) + lambda(2i+k) and r= (2i-j) + mu(i+j-k)

Asked by lekhakarthikeyan | 09 Jan, 2019, 04:45: PM
Expert Answer
begin mathsize 16px style straight i with hat on top minus straight j with hat on top plus straight lambda open parentheses 2 straight i with hat on top plus straight k with hat on top close parentheses space and space stack 2 straight i with hat on top minus straight j with hat on top plus straight mu open parentheses straight i with hat on top plus straight j with hat on top minus straight k with hat on top close parentheses
straight x subscript 1 equals 1 comma straight y subscript 1 equals negative 1 comma space straight z subscript 1 equals 0 comma straight x subscript 2 equals 2 comma straight y subscript 1 equals negative 1 comma space straight z subscript 1 equals 0
straight a subscript 1 equals 2 comma straight b subscript 1 0 comma straight c subscript 1 equals 1 comma straight a subscript 2 equals 1 comma straight b subscript 2 equals 1 comma straight c subscript 2 equals negative 1


Shortest space distance space equals open vertical bar fraction numerator open vertical bar table row cell straight x subscript 2 minus straight x subscript 1 end cell cell straight y subscript 2 minus straight y subscript 1 end cell cell straight z subscript 2 minus straight z subscript 1 end cell row cell straight a subscript 1 end cell cell straight b subscript 1 end cell cell straight c subscript 1 end cell row cell straight a subscript 2 end cell cell straight b subscript 2 end cell cell straight c subscript 2 end cell end table close vertical bar over denominator square root of open parentheses straight b subscript 1 straight c subscript 2 minus straight b subscript 2 straight c subscript 1 close parentheses squared plus open parentheses straight c subscript 1 straight a subscript 2 minus straight c subscript 2 straight a subscript 1 close parentheses squared plus open parentheses straight a subscript 1 straight b subscript 2 minus straight a subscript 2 straight b subscript 1 close parentheses squared end root end fraction close vertical bar
equals open vertical bar fraction numerator begin display style open vertical bar table row cell 2 minus 1 end cell cell negative 1 plus 1 end cell 0 row 2 0 1 row 1 1 cell negative 1 end cell end table close vertical bar end style over denominator begin display style square root of open parentheses 0 minus 1 close parentheses squared plus open parentheses negative 2 minus 1 close parentheses squared plus open parentheses 2 minus 0 close parentheses squared end root end style end fraction close vertical bar equals open vertical bar fraction numerator negative 1 over denominator square root of 14 end fraction close vertical bar equals fraction numerator begin display style 1 end style over denominator begin display style square root of 14 end style end fraction end style
Answered by Sneha shidid | 10 Jan, 2019, 11:19: AM
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