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 $=<-1/sqrt42,-5/sqrt42,4/sqrt42>$

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