Math Problem: Let $A=(1,-11,2), B=(3,-4,1)$, and $C=(-2,1,-1)$. Compute $\angle A B C$, in degrees.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the angle $ \angle ABC $, we first need to find the vectors $ \overrightarrow{AB} $ and $ \overrightarrow{BC} $. ‖ ‖ step 2 ⋮ The vector $ \overrightarrow{AB} $ is found by subtracting coordinates of $ A $ from $ B $: $ \overrightarrow{AB} = B - A = (3 - 1, -4 - (-11), 1 - 2) $. ‖ ‖ step 3 ⋮ Simplifying the coordinates of $ \overrightarrow{AB} $ gives us $ \overrightarrow{AB} = (2, 7, -1) $. ‖ ‖ step 4 ⋮ Similarly, the vector $ \overrightarrow{BC} $ is $ \overrightarrow{BC} = C - B = (-2 - 3, 1 - (-4), -1 - 1) $. ‖ ‖ step 5 ⋮ Simplifying the coordinates of $ \overrightarrow{BC} $ gives us $ \overrightarrow{BC} = (-5, 5, -2) $. ‖ ‖ step 6 ⋮ The angle between two vectors $ \overrightarrow{u} $ and $ \overrightarrow{v} $ can be found using the dot product formula: $ \cos(\theta) = \frac{\overrightarrow{u} \cdot \overrightarrow{v}}{|\overrightarrow{u}| |\overrightarrow{v}|} $. ‖ ‖ step 7 ⋮ Calculate the dot product $ \overrightarrow{AB} \cdot \overrightarrow{BC} = (2)(-5) + (7)(5) + (-1)(-2) $. ‖ ‖ step 8 ⋮ Simplifying the dot product gives us $ \overrightarrow{AB} \cdot \overrightarrow{BC} = -10 + 35 + 2 = 27 $. ‖ ‖ step 9 ⋮ Calculate the magnitudes $ |\overrightarrow{AB}| = \sqrt{2^2 + 7^2 + (-1)^2} $ and $ |\overrightarrow{BC}| = \sqrt{(-5)^2 + 5^2 + (-2)^2} $. ‖ ‖ step 10 ⋮ Simplifying the magnitudes gives us $ |\overrightarrow{AB}| = \sqrt{4 + 49 + 1} = \sqrt{54} $ and $ |\overrightarrow{BC}| = \sqrt{25 + 25 + 4} = \sqrt{54} $. ‖ ‖ step 11 ⋮ Now we can find $ \cos(\theta) = \frac{27}{\sqrt{54} \sqrt{54}} $. ‖ ‖ step 12 ⋮ Simplifying $ \cos(\theta) $ gives us $ \cos(\theta) = \frac{27}{54} = \frac{1}{2} $. ‖ ‖ step 13 ⋮ To find the angle in degrees, we use the inverse cosine function: $ \theta = \cos^{-1}\left(\frac{1}{2}\right) $. ‖ ‖ step 14 ⋮ The value of $ \cos^{-1}\left(\frac{1}{2}\right) $ is $ 60^\circ $. ‖ ∻Answer∻ ⚹ $ \angle ABC = 60^\circ $ ⚹ ∻Key Concept∻ ⚹Angle between two vectors using dot product⚹ ∻Explanation∻ ⚹The angle between two vectors is calculated by taking the inverse cosine of the dot product of the vectors divided by the product of their magnitudes.⚹◊What is the formula to compute the angle between two vectors in three dimensions?⍭ Generate me a similar question◊

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