How would you find the unit vector along the line joining point (2, 4, 4) to point (3, 2, 2)?
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Please see the explanation.
You make a vector in the direction from one point to another by subtracting each starting coordinate from its respective ending coordinate:
Therefore the general form for a unit vector is:
Compute the magnitude
Substitute into the unit vector general form:
Simplify:
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To find the unit vector along the line joining point (2, 4, 4) to point (3, 2, 2), follow these steps:
 Determine the direction vector by subtracting the coordinates of the two points: ( \vec{v} = (3  2, 2  4, 2  4) = (5, 2, 2) ).
 Find the magnitude of the direction vector: ( \vec{v} = \sqrt{(5)^2 + (2)^2 + (2)^2} = \sqrt{25 + 4 + 4} = \sqrt{33} ).
 Divide the direction vector by its magnitude to obtain the unit vector: ( \hat{v} = \frac{\vec{v}}{\vec{v}} = \left( \frac{5}{\sqrt{33}}, \frac{2}{\sqrt{33}}, \frac{2}{\sqrt{33}} \right) ).
Therefore, the unit vector along the line joining the two given points is ( \left( \frac{5}{\sqrt{33}}, \frac{2}{\sqrt{33}}, \frac{2}{\sqrt{33}} \right) ).
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To find the unit vector along the line joining the points (2, 4, 4) and (3, 2, 2), follow these steps:
 Find the vector that represents the direction of the line by subtracting the coordinates of the second point from the coordinates of the first point.
 Calculate the magnitude of the direction vector.
 Divide each component of the direction vector by its magnitude to obtain the unit vector.
Let's denote the points as ( P_1(2, 4, 4) ) and ( P_2(3, 2, 2) ).

The direction vector, ( \mathbf{v} ), is found by subtracting the coordinates of ( P_1 ) from the coordinates of ( P_2 ): [ \mathbf{v} = \langle 3  2, 2  4, 2  4 \rangle = \langle 5, 2, 2 \rangle ]

The magnitude of ( \mathbf{v} ), denoted as (  \mathbf{v}  ), is calculated using the formula: [  \mathbf{v}  = \sqrt{(5)^2 + (2)^2 + (2)^2} = \sqrt{25 + 4 + 4} = \sqrt{33} ]

The unit vector, ( \mathbf{u} ), along the line is obtained by dividing each component of ( \mathbf{v} ) by its magnitude: [ \mathbf{u} = \left\langle \frac{5}{\sqrt{33}}, \frac{2}{\sqrt{33}}, \frac{2}{\sqrt{33}} \right\rangle ]
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When evaluating a onesided 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.
When evaluating a onesided 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.
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