Find the area of parallelogram with adjacent sides a=2i+3j and b=i-4j

Question

Find the area of the parallelogram whose adjacent sides are given by $\vec{a} = 2\hat{i} + 3\hat{j}$ and $\vec{b} = \hat{i} - 4\hat{j}$ .

Solution — Step by Step

The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ equals the magnitude of their cross product:

\text{Area} = |\vec{a} \times \vec{b}|

For 2D vectors (in the $xy$ -plane), we treat them as 3D vectors with $z$ -component = 0.

\vec{a} = 2\hat{i} + 3\hat{j} + 0\hat{k}

\vec{b} = 1\hat{i} - 4\hat{j} + 0\hat{k}

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 1 & -4 & 0 \end{vmatrix}

Expanding along the first row:

= \hat{i}(3 \times 0 - 0 \times (-4)) - \hat{j}(2 \times 0 - 0 \times 1) + \hat{k}(2 \times (-4) - 3 \times 1)

= \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(-8 - 3)

= 0\hat{i} + 0\hat{j} - 11\hat{k}

\vec{a} \times \vec{b} = -11\hat{k}

|\vec{a} \times \vec{b}| = \sqrt{0^2 + 0^2 + (-11)^2} = \sqrt{121} = 11

Area of the parallelogram = 11 square units

Why This Works

The cross product $\vec{a} \times \vec{b}$ produces a vector perpendicular to both $\vec{a}$ and $\vec{b}$ , with magnitude $|\vec{a}||\vec{b}|\sin\theta$ , where $\theta$ is the angle between them. Geometrically, $|\vec{a}||\vec{b}|\sin\theta$ is exactly the base times height of the parallelogram.

For 2D vectors in the $xy$ -plane, the cross product is purely in the $z$ -direction, making the calculation especially clean. The $z$ -component of the cross product ( $a_x b_y - a_y b_x$ ) is all we need: $2 \times (-4) - 3 \times 1 = -8 - 3 = -11$ .

Alternative Method — Direct 2D Formula

For 2D vectors, Area $= |a_x b_y - a_y b_x|$

$= |2 \times (-4) - 3 \times 1| = |-8 - 3| = |-11| = 11$ square units.

This is the magnitude of the $z$ -component of the cross product directly. Faster for 2D problems.

Common Mistake

When computing the determinant expansion for the cross product, students often get the sign wrong for the $\hat{j}$ component — it should be $-\hat{j}(\ldots)$ , not $+\hat{j}(\ldots)$ . The cofactor expansion of a 3×3 determinant alternates in sign: $+$ for $\hat{i}$ , $-$ for $\hat{j}$ , $+$ for $\hat{k}$ . Missing the negative sign on the $\hat{j}$ term is the most common error in cross product calculations.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct 2D Formula

Common Mistake

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