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Cross Product Calculator

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Example
Created on 2024-06-20Asked by Ethan Thompson (Solvelet student)
Compute the cross product of the vectors u=2,3,4 \mathbf{u} = \langle 2, -3, 4 \rangle and v=1,5,2 \mathbf{v} = \langle -1, 5, 2 \rangle .

Solution

Step 1: Identify the vectors u=2,3,4 \mathbf{u} = \langle 2, -3, 4 \rangle and v=1,5,2 \mathbf{v} = \langle -1, 5, 2 \rangle . Step 2: Use the formula for the cross product of two vectors: u×v=ijk 234 152. \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -3 & 4 \ -1 & 5 & 2 \end{vmatrix}. Step 3: Expand the determinant along the top row: u×v=i34 52j24 12+k23 15. \mathbf{u} \times \mathbf{v} = \mathbf{i} \begin{vmatrix} -3 & 4 \ 5 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 4 \ -1 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -3 \ -1 & 5 \end{vmatrix}. Step 4: Compute the determinants: u×v=i((3)(2)(4)(5))j((2)(2)(4)(1))+k((2)(5)(3)(1)). \mathbf{u} \times \mathbf{v} = \mathbf{i} ((-3)(2) - (4)(5)) - \mathbf{j} ((2)(2) - (4)(-1)) + \mathbf{k} ((2)(5) - (-3)(-1)). Step 5: Perform the calculations: u×v=i(620)j(4+4)+k(103). \mathbf{u} \times \mathbf{v} = \mathbf{i} (-6 - 20) - \mathbf{j} (4 + 4) + \mathbf{k} (10 - 3). Step 6: Simplify: u×v=26i8j+13k. \mathbf{u} \times \mathbf{v} = -26\mathbf{i} - 8\mathbf{j} + 13\mathbf{k}. Step 7: Conclusion. The cross product of u \mathbf{u} and v \mathbf{v} is u×v=26,8,13 \mathbf{u} \times \mathbf{v} = \langle -26, -8, 13 \rangle . Solved on Solvelet with Basic AI Model
Some of the related questions asked by Levi Sanchez on Solvelet
1. Calculate the cross product of the vectors u=3,2,5 \mathbf{u} = \langle 3, -2, 5 \rangle and v=1,4,2 \mathbf{v} = \langle 1, 4, -2 \rangle 2. Find a unit vector orthogonal to both i+j \mathbf{i} + \mathbf{j} and 2ij+k 2\mathbf{i} - \mathbf{j} + \mathbf{k} .,
DefinitionA vector product is a function of two vectors whose values ​​are also vectors. Additionally, in the cross product, the vectors are perpendicular to the plane formed by the two vectors. The value of the length product is equal to the area of ​​the parallelogram represented by two vectors also pointing to the right.
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