How do you normalize #(- 2i - j - k)#?

Answer 1

#hatv = -sqrt(2/3)hati - 1/(sqrt6)hatj - 1/(sqrt6)hatk#

Finding a unit vector that points in the same direction as the vector in question is the process of normalizing a vector.

The equation for the normalization of a vector (which I'll call #vecv#) is given by

#(vecv)/(||vecv||)#

where #||vecv||# is the magnitude of vector #vecv#.

The magnitude of #vecv# is

#||vecv|| = sqrt((-2)^2 + (-1)^2 + (-1)^2) = color(red)(ul(sqrt6#

Thus, the unit vector (#hatv#) will be

#hatv = (-2)/(color(red)(sqrt6))hati - 1/(color(red)(sqrt6))hatj - 1/(color(red)(sqrt6))hatk#

#color(blue)(ulbar(|stackrel(" ")(" "hatv = -sqrt(2/3)hati - 1/(sqrt6)hatj - 1/(sqrt6)hatk" ")|)#

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Answer 2

Nora Arzate

To normalize the vector ((-2i - j - k)), calculate the magnitude and divide each component by the magnitude. The normalized vector is (\frac{-2i - j - k}{\sqrt{6}}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.