Q.Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec{a}=3\hat{i}+\hat{j}+4\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$.

Concept: Cross Product Area — the area of a parallelogram formed by two adjacent vectors equals the magnitude of their cross product $|\vec{a}\times\vec{b}|$.

Step 1: Set up and expand $\vec{a}\times\vec{b}$ for $\vec{a}=3\hat{i}+\hat{j}+4\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$:

Step 2: Simplify each component …

The area of a parallelogram whose adjacent sides are the vectors $\vec{a}$ and $\vec{b}$ is not the plain product of their lengths — that only holds when the sides are perpendicular. In general the area is the magnitude of the cross product, $|\vec{a}\times\vec{b}|$, because $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta$, which is exactly base $\times$ height for the parallelogram.

1. Write the vectors in component form

2. Set up the cross-product determinant

3. Expand along the first row

Evaluating each $2\times2$ minor:

• Coefficient of $\hat{i}$: $(1)(1)-(4)(-1)=1+4=5$
• Coefficient of $\hat{j}$: $-\big((3)(1)-(4)(1)\big)=-(3-4)=1$
• Coefficient of $\hat{k}$: $(3)(-1)-(1)(1)=-3-1=-4$

Hence

$\vec{a}\times\vec{b}=5\hat{i}+\hat{j}-4\hat{k}.$ …

Method: Area of a parallelogram from its adjacent side vectors

When the two adjacent sides are given as vectors, the area is the magnitude of their cross product — no angle needed.

Steps

Step 1: Identify the two adjacent side vectors

Use the vectors $\vec{a}$ and $\vec{b}$ lying along the adjacent sides directly (or, from vertices, form them by subtraction).

Step 2: Cross them with a determinant

$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}.$ …

Common Mistakes

Mistake 1: Using $|\vec{a}|\,|\vec{b}|$ (product of lengths) instead of $|\vec{a}\times\vec{b}|$.

Why it's wrong: $|\vec{a}||\vec{b}|$ overcounts unless the sides are perpendicular, because the true area is $|\vec{a}||\vec{b}|\sin\theta$, not $|\vec{a}||\vec{b}|$. Correct approach: compute the cross-product vector and take its magnitude.

Mistake 2: Halving the answer.

Why it's wrong: the $\tfrac{1}{2}$ factor is only for a triangle; a parallelogram is the full cross-product magnitude. Correct approach: report $|\vec{a}\times\vec{b}|$ with no $\tfrac{1}{2}$. …