Question 206228
to get the unit vector you need to get the magnitude of the vector and then divide each measure used by the magnitude.
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in your problem, you have (i,j,k) = (3,1,2)
magnitude = {{{sqrt(i^2 + j^2 + k^2)}}} = {{{sqrt(3^2 + 1^2 + 2^2)}}} = {{{sqrt(9+1+4)}}} = {{{sqrt(14)}}}.
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you take each measure and divide by the magnitude to get ({{{3/sqrt(14)}}},{{{1/sqrt(14)}}},{{{2/sqrt(14)}}})
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you simplify the fraction by multiplying each by {{{sqrt(14)/sqrt(14)}}} to get ({{{3*sqrt(14)/14}}},{{{sqrt(14)/14}}},{{{2*sqrt(14)/14)}}}) which can be simplified further to get ({{{3*sqrt(14)/14}}},{{{sqrt(14)/14}}},{{{(sqrt(14)/7)}}}) although I think I would want to keep all denominators the same.
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your answer will either be ({{{3*sqrt(14)/14}}},{{{sqrt(14)/14}}},{{{2*sqrt(14)/14)}}}) or ({{{3*sqrt(14)/14}}},{{{sqrt(14)/14}}},{{{sqrt(14)/7)}}})
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this is a unit vector because the magnitude is now 1 as shown by the following equation:
magnitude of vector = {{{sqrt((3*sqrt(14)/14)^2 + (sqrt(14)/14)^2 + (2*sqrt(14))/14)^2)}}}
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this becomes {{{sqrt((9*14)/(14^2) + (14)/(14^2) + (4*14)/(14^2))}}} which becomes {{{sqrt(((9*14) + (14) + (4*14))/(14^2))}}} which becomes {{{sqrt((14^2)/(14^2))}}} which becomes {{{1}}}.
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A unit vector in the opposite direction would have the same values only negatives of whatever is there.
In this case, your unit vector in the opposite direction should be ({{{-(3*sqrt(14)/14)}}},{{{-(sqrt(14)/14)}}},{{{-(2*sqrt(14)/14))}}})
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Note that keeping the denominator the same seemed like a good idea especially when verifying that the magnitude is 1.
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