Cross Product Math Example 3

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Example 3

medium
Find ⟨1,2,0⟩×⟨3,0,0⟩\langle 1, 2, 0 \rangle \times \langle 3, 0, 0 \rangle.

Solution

  1. 1
    ii: 2(0)−0(0)=02(0) - 0(0) = 0. jj: 0(3)−1(0)=00(3) - 1(0) = 0. kk: 1(0)−2(3)=−61(0) - 2(3) = -6.
  2. 2
    Result: ⟨0,0,−6⟩\langle 0, 0, -6 \rangle.

Answer

⟨0,0,−6⟩\langle 0, 0, -6 \rangle
When both vectors lie in the xyxy-plane (z-component = 0), their cross product points purely in the zz-direction. The magnitude ∣−6∣|{-6}| equals the area of the parallelogram spanned by the two vectors.

About Cross Product

The cross product of two 3D vectors a=⟨a1,a2,a3⟩\mathbf{a} = \langle a_1, a_2, a_3 \rangle and b=⟨b1,b2,b3⟩\mathbf{b} = \langle b_1, b_2, b_3 \rangle is a new vector a×b\mathbf{a} \times \mathbf{b} that is perpendicular to both a\mathbf{a} and b\mathbf{b}. Its magnitude equals the area of the parallelogram formed by a\mathbf{a} and b\mathbf{b}.

Learn more about Cross Product →

More Cross Product Examples

Example 1 medium

Find [formula].

Example 2 hard

Find [formula].

Example 4 easy

Is [formula]?

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