What is the magnitude of the vector 3i-j+2k in the direction of 30° from the above vector?

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Answer 1 · 2017-08-23T08:33:42

The answer is #=3.12u#

The magnitude of a vector #veca= <a,b,c ># is

#||< a,b,c >||= sqrt(a^2+b^2+c^2)#

Here,

we have

#||< 3,-1,2>|| = sqrt((3)^2+(-1)^2+(2)^2)#

#=sqrt(9+1+4)=sqrt13#

The component at #30^@# is

#=sqrt13cos(30^@)=3.12u#