If \(\vec{a}=\hat{\imath}-2\hat{\jmath}+3\hat{k}\) and \(\vec{b}=3\hat{\imath}+\hat{\jmath}-\hat{k}\), then \((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3
Expanding by the distributive law, the middle terms cancel because \(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\):
\((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2-|\vec{b}|^2\)
\(|\vec{a}|^2 = 1+4+9 = 14\)
\(|\vec{b}|^2 = 9+1+1 = 11\)
\(|\vec{a}|^2-|\vec{b}|^2 = 14-11 = 3\)
Check with components: \(\vec{a}+\vec{b} = 4\hat{\imath}-\hat{\jmath}+2\hat{k}\) and \(\vec{a}-\vec{b} = -2\hat{\imath}-3\hat{\jmath}+4\hat{k}\), so the product is \(-8+3+8 = 3\).