Let \(\vec{a}\) and \(\vec{b}\) be any two vectors.
Match List-I with List-II
| List-I | List-II | ||
| (A) | \((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})\) | (I) | \(4\,(\vec{a}\cdot\vec{b})\) |
| (B) | \(|\vec{a}+\vec{b}|^2-|\vec{a}-\vec{b}|^2\) | (II) | \(-2\,(\vec{a}\times\vec{b})\) |
| (C) | \((\vec{a}+\vec{b})\times(\vec{a}-\vec{b})\) | (III) | \(|\vec{a}|^2-|\vec{b}|^2\) |
| (D) | \(|\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2\) | (IV) | \(2\,(|\vec{a}|^2+|\vec{b}|^2)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
Each product is expanded by the distributive law, using \(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\), \(\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}\) and \(\vec{v}\times\vec{v} = \vec{0}\).
(A) \((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})\)
This matches (III) \(|\vec{a}|^2-|\vec{b}|^2\).
\( = |\vec{a}|^2-\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{a}-|\vec{b}|^2\)
\( = |\vec{a}|^2-|\vec{b}|^2\)
(B) \(|\vec{a}+\vec{b}|^2-|\vec{a}-\vec{b}|^2\)
This matches (I) \(4\,(\vec{a}\cdot\vec{b})\).
\( = (|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b})-(|\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b})\)
\( = 4\,(\vec{a}\cdot\vec{b})\)
(C) \((\vec{a}+\vec{b})\times(\vec{a}-\vec{b})\)
This matches (II) \(-2\,(\vec{a}\times\vec{b})\).
\( = \vec{a}\times\vec{a}-\vec{a}\times\vec{b}+\vec{b}\times\vec{a}-\vec{b}\times\vec{b}\)
\( = \vec{0}-\vec{a}\times\vec{b}-\vec{a}\times\vec{b}-\vec{0}\)
\( = -2\,(\vec{a}\times\vec{b})\)
(D) \(|\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2\)
This matches (IV) \(2\,(|\vec{a}|^2+|\vec{b}|^2)\).
\( = (|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b})+(|\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b})\)
\( = 2\,(|\vec{a}|^2+|\vec{b}|^2)\)