Question

Explain how to find the direction of the cross product $\mathbf{u}\times \mathbf{v}$

Short answer

Question: Determine the direction of the cross product of two vectors $\mathbf{u}$ and $\mathbf{v}$. Answer: To find the direction of the cross product $\mathbf{u}\times \mathbf{v}$, first calculate the cross product using the formula: $$\mathbf{u}\times \mathbf{v}=\left(\begin{array}{c}{u}_2{v}_3-u_3{v}_2\\ u_3{v}_1-u_1{v}_3\\ u_1{v}_2-u_2{v}_1\end{array}\right)$$ Then, find the magnitude of the cross product: $$||\mathbf{u}\times \mathbf{v}||=\sqrt{(u_2{v}_3-u_3{v}_2{)}^2+(u_3{v}_1-u_1{v}_3{)}^2+(u_1{v}_2-u_2{v}_1{)}^2}$$ Finally, divide the cross product vector by its magnitude to find the unit vector representing the direction of $\mathbf{u}\times \mathbf{v}$: $$\hat{\mathbf{w}}=\frac{\mathbf{u}\times \mathbf{v}}{||\mathbf{u}\times \mathbf{v}||}=\left(\begin{array}{c}\frac{u_2{v}_3-u_3{v}_2}{||\mathbf{u}\times \mathbf{v}||}\\ \frac{u_3{v}_1-u_1{v}_3}{||\mathbf{u}\times \mathbf{v}||}\\ \frac{u_1{v}_2-u_2{v}_1}{||\mathbf{u}\times \mathbf{v}||}\end{array}\right)$$

Step-by-step solution

Calculate the cross product $\mathbf{u}\times \mathbf{v}$

Let $\mathbf{u}=\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right)$ and $\mathbf{v}=\left(\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right)$. The cross product is given by: $$\mathbf{u}\times \mathbf{v}=\left(\begin{array}{c}{u}_2{v}_3-u_3{v}_2\\ u_3{v}_1-u_1{v}_3\\ u_1{v}_2-u_2{v}_1\end{array}\right)$$

Find the magnitude of $\mathbf{u}\times \mathbf{v}$

To find the direction, we first need to determine the magnitude of the cross product. This is equal to the length of the vector $\mathbf{u}\times \mathbf{v}$, given by the formula: $$||\mathbf{u}\times \mathbf{v}||=\sqrt{(u_2{v}_3-u_3{v}_2{)}^2+(u_3{v}_1-u_1{v}_3{)}^2+(u_1{v}_2-u_2{v}_1{)}^2}$$

Find the direction of $\mathbf{u}\times \mathbf{v}$ as a unit vector

Finally, to find the direction of the cross product, we need to divide the cross product vector by its magnitude to find the unit vector: $$\hat{\mathbf{w}}=\frac{\mathbf{u}\times \mathbf{v}}{||\mathbf{u}\times \mathbf{v}||}=\left(\begin{array}{c}\frac{u_2{v}_3-u_3{v}_2}{||\mathbf{u}\times \mathbf{v}||}\\ \frac{u_3{v}_1-u_1{v}_3}{||\mathbf{u}\times \mathbf{v}||}\\ \frac{u_1{v}_2-u_2{v}_1}{||\mathbf{u}\times \mathbf{v}||}\end{array}\right)$$ The unit vector $\hat{\mathbf{w}}$ represents the direction of the cross product $\mathbf{u}\times \mathbf{v}$.

Key concepts

Vector Calculus

In the realm of mathematics, vector calculus is an essential tool that helps us to handle problems involving vectors and their properties like direction and magnitude. Vectors are important because they have both magnitude and direction, making them suitable for representing physical quantities such as force and velocity. Vector calculus provides the methods to work with these vectors systematically.

One of the operations that vector calculus allows us to perform is the cross product, which is a way to find a vector perpendicular to two given vectors in three-dimensional space. Its result is another vector, not just a number. The cross product is mathematically defined, but its real-world applications are vast and include physics and engineering, where determining the direction perpendicular to a plane is crucial.

Vector calculus also introduces us to other vector operations, such as the dot product, which yields a scalar rather than a vector. Together, these operations allow us to solve complex spatial problems by manipulating vectors according to set rules. The focus of vector calculus is to extend the capabilities we have with basic algebra to work with the multi-dimensional nature of vectors.

Magnitude

Magnitude is a key concept when dealing with vectors, including their cross products. Essentially, the magnitude of a vector represents its length or size, and is a crucial factor in quantifying its properties.

Calculating the magnitude of a vector involves using the Pythagorean theorem in three dimensions. For a vector $\mathbf{a}=\left(\begin{array}{c}{a}_1{a}_2{a}_3\end{array}\right)$, the magnitude $||\mathbf{a}||$ is given by the equation:

• $||\mathbf{a}||=\sqrt{a_1^2+a_2^2+a_3^2}$

When applying this to the cross product operation, the magnitude tells us how "strong" or "influential" the resulting vector is compared to other vectors in the space. The magnitude of the cross product $\mathbf{u}\times \mathbf{v}$ reflects the area of the parallelogram formed by the two vectors $\mathbf{u}$ and $\mathbf{v}$. This makes it particularly useful in areas like physics, where understanding force fields and line integrals are necessary.

Understanding and calculating the magnitude is vital for determining the intensity or extent of the vector's effect in the context of the given problem.

Unit Vector

A unit vector is an important concept when discussing direction in vector calculus. It is simply a vector with a magnitude of exactly one. This means unit vectors retain the direction of the original vector but have a standardized length. Such vectors provide a simplified way to express direction alone, without size or intensity.

To find a unit vector from any given vector $\mathbf{a}$, you divide the vector by its magnitude. Thus, the unit vector $\hat{\mathbf{a}}$ is calculated as:

• $\hat{\mathbf{a}}=\frac{\mathbf{a}}{||\mathbf{a}||}$

For the cross product $\mathbf{u}\times \mathbf{v}$, finding the unit vector involves dividing the cross product vector by its magnitude, which will point in the direction determined by the right-hand rule. This method ensures you have a clear and normalized representation of direction.

Unit vectors are particularly useful for scaling vectors and ensuring calculations remain manageable without losing directional information, which is crucial when vector operations involve multiple dimensions.