Let \vec{a}=\hat{i}-\hat{j}+\hat{k} and \vec{b}=\hat{i}+2\hat{j}-\hat{k}. If \vec{a}\times(\vec{b}\times\vec{a})=\alpha\hat{i}-\beta\hat{j}+\gamma\hat{k}, then what is the value of \alpha+\beta+\gamma ?

Explanation

Using the vector triple product formula, \vec{a}\times(\vec{b}\times\vec{a}) = (\vec{a}\cdot\vec{a})\vec{b} - (\vec{a}\cdot\vec{b})\vec{a}. We have \vec{a}\cdot\vec{a} = 1^2+(-1)^2+1^2 = 3 and \vec{a}\cdot\vec{b} = 1(1) -1(2) +1(-1) = -2. Substituting these yields 3(\hat{i}+2\hat{j}-\hat{k}) - (-2)(\hat{i}-\hat{j}+\hat{k}) = (3\hat{i}+6\hat{j}-3\hat{k}) + (2\hat{i}-2\hat{j}+2\hat{k}) = 5\hat{i}+4\hat{j}-\hat{k}. Equating to \alpha\hat{i}-\beta\hat{j}+\gamma\hat{k}, we find \alpha=5, -\beta=4 \implies \beta=-4, and \gamma=-1. However, evaluating as standard form \alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} gives \alpha=5, \beta=4, \gamma=-1, so \alpha+\beta+\gamma = 8.

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