Calculate cross product using 3×3 determinant with î, ĵ, k̂.

Cross Product — Find a⃗ × b⃗ Using Determinant Method

Question

Given $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + 4\hat{k}$ , find $\vec{a} \times \vec{b}$ .

Solution — Step by Step

Write the cross product as a determinant with $\hat{i}$ , $\hat{j}$ , $\hat{k}$ in the first row, components of $\vec{a}$ in the second, and components of $\vec{b}$ in the third:

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

This is the standard setup — memorise the row order: unit vectors, first vector, second vector.

We expand using cofactors. Each unit vector multiplies the 2×2 minor formed by deleting its row and column:

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

Notice the alternating signs: $+\hat{i}$ , $-\hat{j}$ , $+\hat{k}$ . This sign pattern trips up many students.

For each minor, use $ad - bc$ :

$\hat{i}$ minor: $(3)(4) - (-1)(-2) = 12 - 2 = 10$
$\hat{j}$ minor: $(2)(4) - (-1)(1) = 8 + 1 = 9$
$\hat{k}$ minor: $(2)(-2) - (3)(1) = -4 - 3 = -7$

Take your time with the negatives here — this is where arithmetic errors creep in.

Substitute back:

\vec{a} \times \vec{b} = 10\hat{i} - 9\hat{j} + (-7)\hat{k}

\boxed{\vec{a} \times \vec{b} = 10\hat{i} - 9\hat{j} - 7\hat{k}}

The cross product must be perpendicular to both $\vec{a}$ and $\vec{b}$ . Verify: $(10)(2) + (-9)(3) + (-7)(-1) = 20 - 27 + 7 = 0$ . ✓

This 10-second check costs nothing and catches errors before they cost marks.

Why This Works

The determinant method is not a trick — it’s a compact encoding of the cross product definition. When we expand the determinant, we’re computing $a_y b_z - a_z b_y$ for the $\hat{i}$ component, $a_z b_x - a_x b_z$ for $\hat{j}$ , and $a_x b_y - a_y b_x$ for $\hat{k}$ . The determinant just gives us a systematic way to remember all six terms without cramming a formula.

Geometrically, $\vec{a} \times \vec{b}$ gives a vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ . Its magnitude equals the area of the parallelogram formed by the two vectors. This is why cross products appear in torque, angular momentum, and area calculations — all topics with direct weightage in JEE Main.

The right-hand rule tells us the direction: curl fingers from $\vec{a}$ toward $\vec{b}$ , and the thumb points in the direction of $\vec{a} \times \vec{b}$ . For numerical problems, the determinant handles direction automatically through the sign.

Alternative Method — Component Formula

If you prefer direct formulas over determinants, use the cyclic component expansion:

\vec{a} \times \vec{b} = (a_y b_z - a_z b_y)\hat{i} - (a_x b_z - a_z b_x)\hat{j} + (a_x b_y - a_y b_x)\hat{k}

Plugging in $\vec{a} = (2, 3, -1)$ and $\vec{b} = (1, -2, 4)$ :

$\hat{i}$ : $(3)(4) - (-1)(-2) = 12 - 2 = 10$
$\hat{j}$ : $-[(2)(4) - (-1)(1)] = -(8 + 1) = -9$
$\hat{k}$ : $(2)(-2) - (3)(1) = -4 - 3 = -7$

Same answer: $10\hat{i} - 9\hat{j} - 7\hat{k}$ .

For JEE, the determinant method is faster under time pressure — one setup, systematic expansion. The component formula works too, but it’s easier to lose the negative sign on the $\hat{j}$ term. Pick one method and drill it so it’s automatic.

Common Mistake

The most common error: forgetting the negative sign on the $\hat{j}$ term during expansion.

Students write $+\hat{j}$ instead of $-\hat{j}$ when expanding the first row, getting $\hat{i} - \hat{j}$ wrong as $\hat{i} + \hat{j}$ . This appeared in JEE Main 2024 and cost full marks for students who set up the determinant correctly but expanded it wrong.

The sign pattern for cofactor expansion along the first row is always $+$ , $-$ , $+$ . Write these signs down explicitly before you expand — don’t rely on memory mid-calculation.

Question

Solution — Step by Step

Why This Works

Alternative Method — Component Formula

Common Mistake

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