Discussion
Let \(\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}\), \(\vec{b} = 2\hat{i} + 4\hat{j} + 8\hat{k}\) and \(\vec{c} = 2\hat{i} + 4\hat{j} + 3\hat{k}\) .Then \((\vec{a} \times \vec{b}) \cdot \vec{c}\) =
Answer: E
The scalar triple product is 0.
\(\vec{b} = 2\hat{i} + 4\hat{j} + 8\hat{k} = 2\vec{a}\), so \(\vec{a}\) and \(\vec{b}\) are parallel and \(\vec{a} \times \vec{b} = \vec{0}\). The dot product of the zero vector with \(\vec{c}\) is 0, whatever \(\vec{c}\) is.
Equivalently, the determinant formed by the three vectors has two proportional rows. The non-zero options come from working out the cross product with an arithmetic slip.
General rule: the scalar triple product is zero whenever two of the three vectors are parallel, or more generally whenever the three are coplanar.
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