Let a and b be non-zero vectors which are not parallel. Which of the…

The correct answer is Option (3) → (A) and (C) only

Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\). Then \(\vec{a}\times\vec{b} = |\vec{a}|\,|\vec{b}|\sin\theta\,\hat{n}\), where \(\hat{n}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\).

A parallelogram with adjacent sides \(\vec{a}\), \(\vec{b}\) has base \(|\vec{a}|\) and height \(|\vec{b}|\sin\theta\). The triangle on the same two sides is half of it.

So the parallelogram area is \(|\vec{a}\times\vec{b}|\), the triangle area is \(\displaystyle\frac{1}{2}|\vec{a}\times\vec{b}|\), and the projection of \(\vec{a}\) on \(\vec{b}\) is \(\displaystyle\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\).

Checking the statements

(A) Parallelogram area is \(|\vec{a}\times\vec{b}|\) — Correct. Base \(|\vec{a}|\) times height \(|\vec{b}|\sin\theta\) is \(|\vec{a}\times\vec{b}|\).

(B) Triangle area is \(|\vec{a}\times\vec{b}|\) — Incorrect. The triangle is half the parallelogram, so its area is \(\displaystyle\frac{1}{2}|\vec{a}\times\vec{b}|\).

(C) \(\vec{a}\times\vec{b}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\) — Correct. Its direction \(\hat{n}\) is perpendicular to both vectors.

(D) Projection of \(\vec{a}\) on \(\vec{b}\) is \(\displaystyle\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|}\) — Incorrect. It is \(\displaystyle\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\); dividing by \(|\vec{a}|\) gives the projection of \(\vec{b}\) on \(\vec{a}\).