Discussion
Let \(\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}\), \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\) .Then \((\vec{a} + \vec{b}).(\vec{a} - \vec{b}) =\)
Answer: C
The value is 8.
The dot product distributes like ordinary multiplication, and the cross terms cancel because \(\vec{a}.\vec{b} = \vec{b}.\vec{a}\).
\[ \begin{aligned} (\vec{a} + \vec{b}).(\vec{a} - \vec{b}) &= |\vec{a}|^{2} - |\vec{b}|^{2} \\ &= (9 + 4 + 4) - (1 + 4 + 4) \\ &= 17 - 9 = 8 \end{aligned} \]
Working directly gives the same: \(\vec{a} + \vec{b} = 4\hat{i} + 4\hat{j}\) and \(\vec{a} - \vec{b} = 2\hat{i} + 4\hat{k}\), whose dot product is 8 + 0 + 0 = 8. The neighbouring values come from an arithmetic slip in one of the squared lengths.
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