Examples on Cross Product
Example 1: Calculate the cross product of the vectors ([Tex]\vec{A}[/Tex] = ❬ 2, -1, 3 ❭) and ([Tex]\vec{B}[/Tex] = ❬ -3, 4, 1 ❭).
Solution:
To calculate cross product of two vectors ( [Tex]\vec{A}[/Tex]) and ( [Tex]\vec{B}[/Tex]), denoted as ([Tex]\vec{A} \times \vec{B}[/Tex]) ,we can use the following formula:
[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}[/Tex]
Given ( [Tex]\vec{A}[/Tex] = ❬ 2, -1, 3 ❭) and ( [Tex]\vec{B}[/Tex] = ❬ -3, 4, 1 ❭), we can substitute these values into the determinant:
[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ -3 & 4 & 1 \end{vmatrix}[/Tex]
Expanding determinant:
[Tex]\vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} – \hat{j} \begin{vmatrix} 2 & 3 \\ -3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix} [/Tex]
Now, compute determinants:
[Tex]\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix}[/Tex] = (-1)(1) – (3)(4) = -1 – 12 = -13
[Tex]\begin{vmatrix} 2 & 3 \\ -3 & 1 \end{vmatrix}[/Tex] = (2)(1) – (3)(-3) = 2 + 9 = 11
[Tex]\begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix}[/Tex] = (2)(4) – (-1)(-3) = 8 – 3 = 5
Finally, substitute the determinants into the cross product:
[Tex]\vec{A} \times \vec{B} = \hat{i}(-13) – \hat{j}(11) + \hat{k}(5)[/Tex]
[Tex]\vec{A} \times \vec{B} = -13\hat{i} – 11\hat{j} + 5\hat{k}[/Tex]
So, cross product of ( [Tex]\vec{A}[/Tex]) and ( [Tex]\vec{B}[/Tex]) is ( [Tex]\vec{A} \times \vec{B}[/Tex] = ❬ -13, -11, 5 ❭).
Example 2: Determine the area of the parallelogram formed by the vectors ([Tex]\vec{A}[/Tex] = ❬ 3, 1, -2 ❭) and ([Tex]\vec{B}[/Tex] = ❬ 2, -1, 3 ❭).
Solution:
To determine the area of the parallelogram formed by two vectors ( [Tex]\vec{A}[/Tex] ) and ( [Tex]\vec{B}[/Tex] ), we can use the magnitude of their cross product ( [Tex]|\vec{A} \times \vec{B}|[/Tex] ). The formula for the magnitude of the cross product is:
[Tex]|\vec{A} \times \vec{B}| = |\vec{A}| \cdot |\vec{B}| \cdot \sin(\theta)[/Tex]
Where ( |[Tex]\vec{A}[/Tex]| ) and ( |[Tex]\vec{B}[/Tex]| ) are the magnitudes of vectors ( [Tex]\vec{A}[/Tex]) and ( [Tex]\vec{B}[/Tex]) respectively, and (θ) is the angle between the two vectors.
Given ( [Tex]\vec{A}[/Tex] = ❬ 3, 1, -2 ❭) and ( [Tex]\vec{B}[/Tex] = ❬ 2, -1, 3 ❭), we can compute their magnitudes:
[Tex]|\vec{A}| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{9 + 1 + 4} [/Tex]= √14
[Tex]|\vec{B}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9}[/Tex] = √14
Next, we calculate the cross product \( \vec{A} \times \vec{B} \) and its magnitude:
[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -1 & 3 \end{vmatrix}[/Tex]
= [Tex]\hat{i} \begin{vmatrix} 1 & -2 \\ -1 & 3 \end{vmatrix} – \hat{j} \begin{vmatrix} 3 & -2 \\ 2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 2 & -1 \end{vmatrix}[/Tex]
[Tex]= \hat{i}(1 \cdot 3 – (-2) \cdot (-1)) – \hat{j}(3 \cdot 3 – (-2) \cdot 2) + \hat{k}(3 \cdot (-1) – 1 \cdot 2) [/Tex]
[Tex]= \hat{i}(3 – 2) – \hat{j}(9 + 4) + \hat{k}(-3 – 2) [/Tex]
[Tex]= \hat{i} + \hat{j} – 5\hat{k} [/Tex]
The magnitude of ( [Tex]\vec{A} \times \vec{B}[/Tex] ) is:
[Tex]|\vec{A} \times \vec{B}| = \sqrt{1^2 + 1^2 + (-5)^2} = \sqrt{1 + 1 + 25}[/Tex] = √27
Now, we can calculate the area of the parallelogram:
Area = |[Tex]\vec{A} \times \vec{B}[/Tex]| = √27
Therefore, the area of the parallelogram formed by the vectors ( [Tex]\vec{A}[/Tex]) and ( [Tex]\vec{B}[/Tex] ) is √27 square units.
Example 3: Given that the cross product of two vectors ([Tex]\vec{A} \times \vec{B}[/Tex]) yields a vector ([Tex]\vec{C}[/Tex] = ❬ 4, 5, -3 ❭), if the magnitude of ([Tex]\vec{A}[/Tex]) is (|[Tex]\vec{A}[/Tex]| = 2) and the magnitude of ([Tex]\vec{B}[/Tex]) is (|[Tex]\vec{B}[/Tex]| = 3), find the angle between ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]).
Solution:
To find the angle between two vectors ( [Tex]\vec{A}[/Tex] ) and ( [Tex]\vec{B}[/Tex]) given their magnitudes and the cross product ( [Tex]\vec{C} = \vec{A} \times \vec{B} [/Tex]), we can use the formula for the magnitude of the cross product:
[Tex]|\vec{A} \times \vec{B}| = |\vec{A}| \cdot |\vec{B}| \cdot \sin(\theta)[/Tex]
Given that ( |[Tex]\vec{A}[/Tex]| = 2 ), ( [Tex]|\vec{B}|[/Tex] = 3 ), and ( [Tex]\vec{C}[/Tex] = ❬ 4, 5, -3 ❭), we can calculate the magnitude of ( [Tex]\vec{C}[/Tex]):
[Tex]|\vec{C}| = \sqrt{4^2 + 5^2 + (-3)^2} = \sqrt{16 + 25 + 9}[/Tex] = √50
Now, we can substitute the given values into the formula for the magnitude of the cross product:
[Tex]|\vec{A} \times \vec{B}| = |\vec{A}| \cdot |\vec{B}| \cdot \sin(\theta) [/Tex]
√50 = 2 · 3 · sin(θ)
√50 = 6 · sin(θ)
To find (θ), we divide both sides by 6:
[Tex]\sin(\theta) = \frac{\sqrt{50}}{6}[/Tex]
Now, we can find ( θ ) by taking the inverse sine (arcsine) of both sides:
[Tex]\theta = \arcsin\left(\frac{\sqrt{50}}{6}\right) [/Tex]
Using a calculator, we find:
[Tex]\theta \approx \arcsin\left(\frac{\sqrt{50}}{6}\right) \approx 57.19^\circ[/Tex]
Therefore, the angle between vectors ( [Tex]\vec{A}[/Tex] ) and ( [Tex]\vec{B}[/Tex] ) is approximately ( 57.19°).
Cross Product
Cross product or vector product is a binary operation on two vectors in a three-dimensional oriented Euclidean vector space. Cross product, also called the vector cross product, is a mathematical operation performed on two vectors in three-dimensional space.
In this article, we will understand the meaning of cross product, its definition, the formula of the cross product, the cross product of perpendicular vectors, the cross product of parallel vectors, the right-hand rule cross product and the properties of the cross product.
Table of Content
- What is Cross Product?
- Formula of Cross Product
- Cross Product of Perpendicular Vectors
- Cross Product of Parallel Vectors
- Right-Hand Rule Cross Product
- Matrix Representation of Cross Product
- Triple Cross Product
- Cross Product Properties
- Application of Cross Product