Given #A = i(x+2y) - j(y+3z) +k (3x-y)#, how do you determine a unit vector parallel to #A# at point #P(1,-1,2)#?

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Answer 1 · 2017-07-15T09:04:21

The unit vector is #=<-1/sqrt42,-5/sqrt42,4/sqrt42>#

The vector is

#vec(A)(x,y,z)=veci(x+2y)+vecj(-y-3z)+veck(3x-y)#

At the point , #P=(1,-1,2)#

#vecA(1,-1,2)=veci(-1)+vecj(-5)+veck(4)#

The unit vector in the direction of #vecA(1,-1,2)# is

#(vecA(1,-1,2))/(||vecA(1,-1,2)||)#

The modulus of #vec(A(1,-1,2))# is

#||vecA(1,-1,2)||=| |veci(-1)+vecj(-5)+veck(4)||#

#=sqrt(1+25+16)#

#=sqrt42#

The unit vector is #=<-1/sqrt42,-5/sqrt42,4/sqrt42>#