Q.Show that area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\dfrac{|\vec{a}\times\vec{b}|}{2}$. Also find the area of the parallelogram whose diagonals are $2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}-\hat{k}$.

If the diagonals of a parallelogram are $\vec{a}$ and $\vec{b}$, its area is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$.

Let the adjacent sides be $\vec{p}$ and $\vec{q}$, so the diagonals are $\vec{a}=\vec{p}+\vec{q}$ and $\vec{b}=\vec{p}-\vec{q}$. Then

using $\vec{p}\times\vec{p}=\vec{q}\times\vec{q}=\vec{0}$ and $\vec{q}\times\vec{p}=-\vec{p}\times\vec{q}$. The parallelogram's area is $|\vec{p}\times\vec{q}|=\dfrac{1}{2}|\vec{a}\times\vec{b}|$.

For $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$: …

Idea

The diagonals of a parallelogram are the sum and difference of its two adjacent sides. Calling the sides $\vec{p}$ and $\vec{q}$, the diagonals are $\vec{p}+\vec{q}$ and $\vec{p}-\vec{q}$. Since the area is $|\vec{p}\times\vec{q}|$, we just express that cross product through the diagonals.

Proving the formula

Let $\vec{a}=\vec{p}+\vec{q}$ and $\vec{b}=\vec{p}-\vec{q}$. Expand distributively:

Now $\vec{p}\times\vec{p}=\vec{0}$, $\vec{q}\times\vec{q}=\vec{0}$, and $\vec{q}\times\vec{p}=-\vec{p}\times\vec{q}$, so

Taking magnitudes,

Applying to the given diagonals

$\vec{a}=2\hat{i}-\hat{j}+\hat{k},\ \vec{b}=\hat{i}+3\hat{j}-\hat{k}$. …

Method: Area of a Parallelogram from Its Diagonals

Apply this when a parallelogram is described through its two diagonals rather than its sides.

Steps

Step 1: Express the sides through the diagonals

If the diagonals are $\vec{a}$ and $\vec{b}$ and the sides are $\vec{p},\vec{q}$, then the diagonals are the sum and difference of the sides: $\vec{a}=\vec{p}+\vec{q}$, $\vec{b}=\vec{p}-\vec{q}$. This structural fact is the whole key.

Step 2: Relate the diagonal cross product to the side cross product

Expand $\vec{a}\times\vec{b}=(\vec{p}+\vec{q})\times(\vec{p}-\vec{q})$. The like terms vanish ($\vec{p}\times\vec{p}=\vec{q}\times\vec{q}=\vec{0}$) and the cross terms combine to $-2(\vec{p}\times\vec{q})$. …

Common Mistakes

Mistake 1: Using $|\vec{a}\times\vec{b}|$ instead of $\tfrac12|\vec{a}\times\vec{b}|$

Why it's wrong: here $\vec{a},\vec{b}$ are the diagonals, and $\vec{a}\times\vec{b}=-2(\vec{p}\times\vec{q})$, so its magnitude is twice the area. Correct approach: halve it — the area is $\tfrac12|\vec{a}\times\vec{b}|$.

Mistake 2: Treating $\vec{a},\vec{b}$ as sides rather than diagonals …