Let \(|\vec{a}| = 4\), \(|\vec{b}| = 3\) and the angle between \(\vec{a}\) and \(\vec{b}\) be \(\displaystyle\frac{\pi}{6}\). The area of the parallelogram with adjacent sides \(\vec{a}+2\vec{b}\) and \(2\vec{a}-\vec{b}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30
The area is the magnitude of the vector product of the adjacent sides. Expanding by the distributive law, with \(\vec{a}\times\vec{a} = \vec{b}\times\vec{b} = \vec{0}\) and \(\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}\):
\((\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b}) = 2\,\vec{a}\times\vec{a}-\vec{a}\times\vec{b}+4\,\vec{b}\times\vec{a}-2\,\vec{b}\times\vec{b}\)
\( = \vec{0}-\vec{a}\times\vec{b}-4\,\vec{a}\times\vec{b}-\vec{0}\)
\( = -5\,(\vec{a}\times\vec{b})\)
Now
\( \displaystyle |\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\frac{\pi}{6} = 4\times3\times\frac{1}{2} = 6 \)
\(\text{Area} = 5\times6 = 30\)
Why the other options are incorrect
Option (2) 6 — Incorrect. This is \(|\vec{a}\times\vec{b}|\), the area of the parallelogram on \(\vec{a}\) and \(\vec{b}\) themselves; the factor 5 is missed.
Option (3) 18 — Incorrect. This takes \(4\,\vec{b}\times\vec{a}\) as \(+4\,\vec{a}\times\vec{b}\), giving \(3\,(\vec{a}\times\vec{b})\); the order of a vector product cannot be changed without a minus sign.
Option (4) \(30\sqrt{3}\) — Incorrect. This uses \(\cos\displaystyle\frac{\pi}{6}\) in place of \(\sin\displaystyle\frac{\pi}{6}\).