Let \(\vec{a}\) and \(\vec{b}\) be vectors with \(|\vec{a}| = 5\), \(|\vec{b}| = 8\) and \(|\vec{a}-\vec{b}| = 7\). The area of the triangle with adjacent sides \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(10\sqrt{3}\)
Squaring the given magnitude gives \(\vec{a}\cdot\vec{b}\):
\(|\vec{a}-\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b}\)
\(49 = 25+64-2\,\vec{a}\cdot\vec{b}\)
\(\vec{a}\cdot\vec{b} = 20\)
Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\):
\( \displaystyle \cos\theta = \frac{20}{5\times8} = \frac{1}{2} \)
\( \displaystyle \theta = \frac{\pi}{3}, \quad \sin\theta = \frac{\sqrt{3}}{2} \)
The area of the triangle is half the magnitude of the vector product:
\( \displaystyle \frac{1}{2}|\vec{a}\times\vec{b}| = \frac{1}{2}\times5\times8\times\frac{\sqrt{3}}{2} \)
\( = 10\sqrt{3}\)
Why the other options are incorrect
Option (1) \(20\sqrt{3}\) — Incorrect. This is \(|\vec{a}\times\vec{b}|\), the area of the parallelogram; the triangle is half of it.
Option (3) 10 — Incorrect. This uses \(\sin\theta = \displaystyle\frac{1}{2}\) (the value of \(\cos\theta\)), but \(\sin\displaystyle\frac{\pi}{3} = \frac{\sqrt{3}}{2}\).
Option (4) 20 — Incorrect. This is \(\displaystyle\frac{1}{2}\times5\times8\), which takes \(\sin\theta = 1\) (a right angle), but \(\sin\theta = \displaystyle\frac{\sqrt{3}}{2}\).