Chapter 12: Problem 11
In Exercises \(9-14,\) sketch the coordinate axes and then include the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u} \times \mathbf{v}\) as vectors starting at the origin. $$ \mathbf{u}=\mathbf{i}-\mathbf{k}, \quad \mathbf{v}=\mathbf{j}+\mathbf{k} $$
Short Answer
Expert verified
Plot \( \mathbf{u} = \langle 1, 0, -1 \rangle \), \( \mathbf{v} = \langle 0, 1, 1 \rangle \), and \( \mathbf{u} \times \mathbf{v} = \langle 1, -1, 1 \rangle \).
Step by step solution
01
Understanding Vectors
First, we need to understand the given vectors \( \mathbf{u} \) and \( \mathbf{v} \). The vector \( \mathbf{u} = \mathbf{i} - \mathbf{k} \) can be written as \( \langle 1, 0, -1 \rangle \) indicating it points 1 unit in the positive x-direction and 1 unit in the negative z-direction. Similarly, the vector \( \mathbf{v} = \mathbf{j} + \mathbf{k} \) translates to \( \langle 0, 1, 1 \rangle \) indicating it points 1 unit in the positive y-direction and 1 unit in the positive z-direction.
02
Sketching Coordinate Axes
Draw the 3D coordinate system with the x, y, and z-axes. The positive x-axis points forward, the positive y-axis points to the right, and the positive z-axis points upward.
03
Plotting Vector \( \mathbf{u} \)
Place the tail of \( \mathbf{u} \) at the origin \( (0, 0, 0) \). From the origin, mark a point at \( (1, 0, -1) \) in the coordinate space and draw an arrow from the origin to this point, representing the vector \( \mathbf{u} \).
04
Plotting Vector \( \mathbf{v} \)
Similarly, start at the origin \( (0, 0, 0) \). From here, go to the point \( (0, 1, 1) \) in the coordinate system to position the vector \( \mathbf{v} \). Draw an arrow from the origin to this point for the vector \( \mathbf{v} \).
05
Calculating Cross Product \( \mathbf{u} \times \mathbf{v} \)
To find \( \mathbf{u} \times \mathbf{v} \), use the determinant method for cross products: \[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & -1 \ 0 & 1 & 1 \end{vmatrix} = (0 - (-1))\mathbf{i} - (1 - 0)\mathbf{j} + (1 - 0)\mathbf{k} = \mathbf{i} - \mathbf{j} + \mathbf{k} \]. So, \( \mathbf{u} \times \mathbf{v} = \langle 1, -1, 1 \rangle \).
06
Plotting Cross Product \( \mathbf{u} \times \mathbf{v} \)
Plot the vector \( \mathbf{u} \times \mathbf{v} \) starting from the origin and pointing to the coordinate \( (1, -1, 1) \). This shows the result of the cross product as the vector perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross Product
The cross product is a crucial operation in vector calculus that yields a vector perpendicular to two given vectors in a three-dimensional space. When calculating the cross product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \), symbolized as \( \mathbf{u} \times \mathbf{v} \), it results in a new vector that is orthogonal to both \( \mathbf{u} \) and \( \mathbf{v} \). This process is particularly important in physics and engineering for finding vectors that represent rotational forces and areas perpendicular to surfaces.
The calculation involves a determinant method. With vectors \( \mathbf{u} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{v} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \mathbf{u} \times \mathbf{v} \) is given by the determinant:
Understanding the cross product is essential for applications that need to consider orientations and angles between vectors.
The calculation involves a determinant method. With vectors \( \mathbf{u} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{v} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \mathbf{u} \times \mathbf{v} \) is given by the determinant:
- \( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \).
- This expands to \((a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}\).
Understanding the cross product is essential for applications that need to consider orientations and angles between vectors.
3D Coordinate System
In mathematics and physics, the 3D coordinate system is a framework used to define the position of points in three-dimensional space. The space is defined by three perpendicular axes, named the x-axis, y-axis, and z-axis, each representing one of the three dimensions.
Visualizing this coordinate system can help us comprehend spatial relationships between vectors and their components. In the 3D coordinate system:
Visualizing this coordinate system can help us comprehend spatial relationships between vectors and their components. In the 3D coordinate system:
- The x-axis usually points forward or towards the observer.
- The y-axis extends to the right.
- The z-axis proceeds upwards.
- Vector \( \mathbf{u} = \mathbf{i} - \mathbf{k} \), translates to \( \langle 1, 0, -1 \rangle \), indicating movement in the x-axis and negative z-axis.
- Vector \( \mathbf{v} = \mathbf{j} + \mathbf{k} \), translates to \( \langle 0, 1, 1 \rangle \), moving along the y-axis and in the positive z-axis.
Vector Representation
Vectors are mathematical entities with both magnitude and direction. They are often represented as arrows in coordinate systems and can be described using unit vectors. In three-dimensional space, vectors are commonly expressed in terms of the unit vectors \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \), which represent directions along the x, y, and z axes respectively.
The representation as \( \mathbf{u} = \langle a, b, c \rangle \) provides a clear view of a vector's orientation and magnitude in 3D space.
The representation as \( \mathbf{u} = \langle a, b, c \rangle \) provides a clear view of a vector's orientation and magnitude in 3D space.
- For \( \mathbf{u} \), each component \( a, b, \) and \( c \) specifies how far the vector extends in the direction of its respective axis.
- Vector \( \mathbf{u} = \mathbf{i} - \mathbf{k} \) is denoted as \( \langle 1, 0, -1 \rangle \), pointing along the x-axis and opposite to the z-direction.
- Similarly, vector \( \mathbf{v} = \mathbf{j} + \mathbf{k} \) translates to \( \langle 0, 1, 1 \rangle \), combining directions in both the y and z axes.