Let $$ \vec{x}=\left[\begin{array}{l}2 \ 1 \ 2\end{array} ight] $$ and $$ \vec{y}=\left[\begin{array}{c}-2 \ 5 \ 5\end{array} ight] $$. Then $$ \vec{x} \cdot \vec{y}=\square $$

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Let $$ \vec{x}=\left[\begin{array}{l}2 \ 1 \ 2\end{array}ight] $$ and $$ \vec{y}=\left[\begin{array}{c}-2 \ 5 \ 5\end{array}ight] $$. Then $$ \vec{x} \cdot \vec{y}=\square $$

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To find the dot product of vectors x and y, we multiply the corresponding components of the vectors and then sum the results.

For x = [2, 1, 2] and y = [-2, 5, 5]:
  Dot product = (2 × -2) + (1 × 5) + (2 × 5)
         = (-4) + 5 + 10
         = 11

So, x ⋅ y = 11.
The dot product of vectors x and y is 11.

Let \( \vec{x}=\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right] \) and \( \vec{y}=\left[\begin{array}{c}-2 \\ 5 \\ 5\end{array}\right] \). Then \( \vec{x} \cdot \vec{y}=\square \)

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