# What is the angle between #<5 , 2 , -1 > # and # < 1, -3 , 1 > #?

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To find the angle between two vectors, you can use the dot product formula:

[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| , |\mathbf{b}| \cos(\theta) ]

Where:

- ( \mathbf{a} ) and ( \mathbf{b} ) are the vectors,
- ( |\mathbf{a}| ) and ( |\mathbf{b}| ) are the magnitudes of the vectors, and
- ( \theta ) is the angle between the vectors.

Given vectors ( \mathbf{a} = \langle 5, 2, -1 \rangle ) and ( \mathbf{b} = \langle 1, -3, 1 \rangle ), we can calculate their dot product:

[ \mathbf{a} \cdot \mathbf{b} = (5)(1) + (2)(-3) + (-1)(1) ]

[ \mathbf{a} \cdot \mathbf{b} = 5 - 6 - 1 ]

[ \mathbf{a} \cdot \mathbf{b} = -2 ]

Next, we need to find the magnitudes of the vectors:

[ |\mathbf{a}| = \sqrt{5^2 + 2^2 + (-1)^2} = \sqrt{30} ]

[ |\mathbf{b}| = \sqrt{1^2 + (-3)^2 + 1^2} = \sqrt{11} ]

Now, substitute the dot product and magnitudes into the formula:

[ -2 = \sqrt{30} \cdot \sqrt{11} \cdot \cos(\theta) ]

[ \cos(\theta) = \frac{-2}{\sqrt{30} \cdot \sqrt{11}} ]

[ \cos(\theta) = \frac{-2}{\sqrt{330}} ]

Finally, find the angle ( \theta ) by taking the inverse cosine:

[ \theta = \arccos\left(\frac{-2}{\sqrt{330}}\right) ]

Calculate the value of ( \theta ), which will give you the angle between the two vectors.

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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.

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.

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.

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