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VectorsLinear algebraCollegeabout 0:55

Find the cross product of ⟨-2, 3, 1⟩ and ⟨4, 0, -5⟩: animated lesson

A narrated, step-by-step vectors video lesson with captions. Final answer: u × v = ⟨-15, -6, -12⟩.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.4 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    Find the cross product of ⟨-2, 3, 1⟩ and ⟨4, 0, -5⟩.
  2. 2. Steps, part 1
    1. u×v=∣ijk−23140−5∣u \times v = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ -2 & 3 & 1 \\ 4 & 0 & -5 \end{vmatrix}
      Write the determinant with unit vectors i, j, k in the first row
    2. (3)(−5)−(1)(0)=−15(3)(-5) - (1)(0) = -15
      i component
    3. −((−2)(−5)−(1)(4))=−6-\big((-2)(-5) - (1)(4)\big) = -6
      j component (note the minus sign in the expansion)
    4. (−2)(0)−(3)(4)=−12(-2)(0) - (3)(4) = -12
      k component
  3. 3. Steps, part 2
    1. ∣u×v∣=405=95≈20.1246118|u \times v| = \sqrt{405} = 9\sqrt{5} \approx 20.1246118
      Magnitude = area of the parallelogram spanned by u and v
    2. (u×v)u=0,(u×v)v=0(u\times v) u = 0,\quad (u\times v) v = 0
      Check orthogonality
  4. 4. The answer
    u×v=⟨−15,−6,−12⟩u \times v = \langle -15, -6, -12 \rangle

    u×v = ⟨-15, -6, -12⟩

  5. 5. Check
    • The result is perpendicular to both input vectors (both dot products are 0).

Transcript

The problem. Here is the problem: Find the cross product of ⟨-2, 3, 1⟩ and ⟨4, 0, negative 5⟩. Cross product of two 3D vectors.

Steps, part 1. Write the determinant with unit vectors i, j, k in the first row. i component. j component (note the minus sign in the expansion). k component.

Steps, part 2. Magnitude equals area of the parallelogram spanned by u and v. Check orthogonality.

The answer. So the answer is u times v equals ⟨-15, negative 6, negative 12⟩. The result is perpendicular to both u and v; its length 9the square root of 5 approximately 20.1246118 is the area of the parallelogram they span (the triangle has half that area).

Check. The result is perpendicular to both input vectors (both dot products are 0).

About vectors

Magnitude, unit vectors, dot products and angles between vectors.

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Find the cross product of ⟨-2, 3, 1⟩ and ⟨4, 0, -5⟩ (Lesson)