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How do you normalize # (2i + 3j – 7k) #?

Answer 1

Nathan Ashcroft

#hat v# = #(2/sqrt62)hat i+(3/sqrt62)hat j-(7/sqrt62)hat k#

Normalizing vector #v=(2i+3j–7k)# means finding a unit vector in the direction given by vector #v#.

For a vector #u=ahat i+bhat j+chat k#, this is represented by #hat u#, and is equal to #u/|u|#,

where #|u|# is absolute value of #u# and is given by #sqrt(a^2+b^2+c^2)#.

As a result, in the example

#hat v=(2i+3j–7k)/sqrt(2^2+3^3+(-7)^2)# or

#(2i+3j–7k)/sqrt62# i.e.

#(2/sqrt62)hat i+(3/sqrt62)hat j-(7/sqrt62)hat k#

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Answer 2

Adam Arnold

To normalize the vector (2i + 3j - 7k), you first calculate its magnitude, then divide each component by the magnitude.

Magnitude: √(2^2 + 3^2 + (-7)^2) = √(4 + 9 + 49) = √62

Normalized vector: (2/√62)i + (3/√62)j + (-7/√62)k

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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.