Calculating the Cross Product with a Determinant

Setting up the 3x3 determinant with i, j, k in the top row and the two vectors beneath, then expanding it column by column - including the minus sign on the j term that is so easy to forget.

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Calculating the Cross Product with a Determinant — Moosa Academy

There is no need to memorise a separate formula for the cross product. Set the two vectors under the unit vectors i, j, k and evaluate it exactly like a 3×3 determinant.

Theorem Setting up the determinant
\vec{U} \times \vec{V} = \begin{vmatrix} i & j & k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}

Row one holds the unit vectors, row two the components of \vec{U} , and row three the components of \vec{V} . The order of the rows matters — swapping \vec{U} and \vec{V} reverses the answer.

Concept Cover a column to get each part
i j k the 2×2 block gives the i part
To find the i part, cover the i column and the top row. The four numbers left form a 2×2 determinant. Repeat for j and for k .

The signs alternate: +i, then −j, then +k. Forgetting the minus in front of j is the single most common error here.

Example U = (3, −2, 1) and V = (3, 3, 1)
The i part — cover column i:
(-2)(1) - (1)(3) = -2 - 3 = -5
The j part — cover column j, and take the negative:
-[(3)(1) - (1)(3)] = -[3 - 3] = 0
The k part — cover column k:
(3)(3) - (-2)(3) = 9 + 6 = 15
⟹ ⟨−5, 0, 15⟩

A zero component is perfectly normal — it simply means the result has no extent in that direction.

Example U = (4, 2, −1) and V = (5, 1, 4)
The i part: (2)(4) - (-1)(1) = 8 + 1 = 9
The j part: -[(4)(4) - (-1)(5)] = -[16 + 5] = -21
The k part: (4)(1) - (2)(5) = 4 - 10 = -6
⟹ ⟨9, −21, −6⟩

Notice how the negative components feed through: -(-1)(1) becomes +1 , and the minus outside the j bracket flips the whole result.

Note The procedure in five steps
1. Build the matrix: row one i, j, k ; row two \vec{U} ; row three \vec{V} .
2. Cover the i column and take the 2×2 determinant — this is the i part, positive.
3. Cover the j column, take the determinant, then negate it.
4. Cover the k column and take the determinant — positive again.
5. Each 2×2 determinant is the main diagonal minus the other diagonal.
Summary
  1. A cross product is evaluated like a 3×3 determinant.
  2. Row one is i, j, k; row two is U; row three is V.
  3. Cover a column to get that component as a 2×2 determinant.
  4. The signs run +i, −j, +k — never omit the minus on j.
  5. (3, −2, 1) × (3, 3, 1) = ⟨−5, 0, 15⟩.
231. Arithmetic Sequences and Series
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