What is the unit vector parallel to A at point P(1,-1,2)?

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SUMMARY

The unit vector parallel to the vector A at point P(1,-1,2) is calculated by first substituting the coordinates into the vector expression A = x(1+2(-1)) - y(-1+3(2)) + z(3(1)-(-1)). This results in A = -1i - 5j + 4k. The magnitude of A is determined using the formula √((-1)² + (-5)² + (4)²), which equals √42. The unit vector is then obtained by dividing A by its magnitude, resulting in the final unit vector.

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How can I calculate the following?
Given A=[tex]\hat{x}[/tex](x+2y)-[tex]\hat{y}[/tex](y+3z)+[tex]\hat{z}[/tex](3x-y)
Determine a unit vector parallel to A at point P(1,-1,2).
 
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It's straight forward substitution. Find the vector A at the given point and then calculate the unit vector by scaling A by the inverse of its length (thus making the new vector's length one).
 
I agree. By substitution A=-1i-5j+4z, The magnitude of A = Sqrt((-1)2+(-5)2+(4)2). Divide the magnitude into A to get the unit vector.
 

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