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MIT 18.02 Multivariable Calculus, Fall 2007 10 Online
OpenStudy (anonymous):

how would we find the unit vector in the direction of v= i+j?

OpenStudy (owlfred):

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OpenStudy (anonymous):

The vector i=<1,0> and j=<0,1> so the i+j=<1+0,0+1>=<1,1>. The length of this vector is easy: |i+j|=\[\sqrt{2}\] to make the vector i+j=<1,1> a unit vector we rescale it by it's length (i.e. divide i+j by its length) , v=(i+j)/(|i+j|) thus we have v=\[1/\sqrt{2} <1,1>\] or \[<1/\sqrt{2}, 1/\sqrt{2}>\] If you check the length of this vector v, you see it indeed does have length =1. It is parallel to the vector i+j because it's components are proportional to the components of i+j=<1,1>.

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