Chapter 12: Problem 3
In Exercises \(1-8,\) find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u} .\) $$ \mathbf{u}=2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}, \quad \mathbf{v}=-\mathbf{i}+\mathbf{j}-2 \mathbf{k} $$
Short Answer
Expert verified
The length of both \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \) is 4; directions are along \( \mathbf{k} \) and \(-\mathbf{k} \) respectively.
Step by step solution
01
Compute \( \mathbf{u} \times \mathbf{v} \)
To find the vector cross product \( \mathbf{u} \times \mathbf{v} \), use the determinant formula for cross products:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -2 & 4 \ -1 & 1 & -2 \end{vmatrix}\]This simplifies to:\[= \mathbf{i}((-2)(-2) - (4)(1)) - \mathbf{j}((2)(-2) - (4)(-1)) + \mathbf{k}((2)(1) - (-2)(-1))\]\[= \mathbf{i}(4 - 4) - \mathbf{j}(-4 + 4) + \mathbf{k}(2 + 2)\]\[= 0\mathbf{i} + 0\mathbf{j} + 4\mathbf{k}\]Thus, \( \mathbf{u} \times \mathbf{v} = 4\mathbf{k} \).
02
Compute \( \mathbf{v} \times \mathbf{u} \)
The cross product \( \mathbf{v} \times \mathbf{u} \) is found by reversing the order of the determinant:\[\mathbf{v} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 1 & -2 \ 2 & -2 & 4 \end{vmatrix}\]This simplifies to:\[= \mathbf{i}((1)(4) - (-2)(-2)) - \mathbf{j}((-1)(4) - (-2)(2)) + \mathbf{k}((-1)(-2) - (1)(2))\]\[= \mathbf{i}(4 - 4) - \mathbf{j}(-4 + 4) + \mathbf{k}(2 - 2)\]\[= 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}\]Thus, \( \mathbf{v} \times \mathbf{u} = -4\mathbf{k} \).
03
Calculate the Length of \( \mathbf{u} \times \mathbf{v} \)
The length of the cross product \( \mathbf{u} \times \mathbf{v} = 4\mathbf{k} \) is given by the magnitude of the vector:\[|\mathbf{u} \times \mathbf{v}| = \sqrt{0^2 + 0^2 + 4^2} = \sqrt{16} = 4\].So, the length of \( \mathbf{u} \times \mathbf{v} \) is 4.
04
Calculate the Length of \( \mathbf{v} \times \mathbf{u} \)
The length of the cross product \( \mathbf{v} \times \mathbf{u} = -4\mathbf{k} \) is also calculated by taking the magnitude:\[|\mathbf{v} \times \mathbf{u}| = \sqrt{0^2 + 0^2 + (-4)^2} = \sqrt{16} = 4\].Therefore, the length of \( \mathbf{v} \times \mathbf{u} \) is 4.
05
Identify the Direction of \( \mathbf{u} \times \mathbf{v} \)
The direction of a vector \( 4\mathbf{k} \) is the same as the positive direction of the \( \mathbf{k} \) axis.
06
Identify the Direction of \( \mathbf{v} \times \mathbf{u} \)
The direction of a vector \( -4\mathbf{k} \) is the same as the negative direction of the \( \mathbf{k} \) axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
Vector magnitude is a fundamental concept in vector mathematics. It represents the length of a vector. To find the magnitude of a vector, we use the formula:
\[|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]where \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}. This formula is derived from the Pythagorean theorem.
To understand this better, imagine you are finding the length of a line extending from the origin to a certain point in 3-dimensional space. The coordinates of this point determine the vector, and the vector's magnitude is the straight-line distance from the origin to this point.
\[|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]where \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}. This formula is derived from the Pythagorean theorem.
To understand this better, imagine you are finding the length of a line extending from the origin to a certain point in 3-dimensional space. The coordinates of this point determine the vector, and the vector's magnitude is the straight-line distance from the origin to this point.
- A vector with zero magnitude is a zero vector, having no direction.
- If a vector's components are proportional, the vectors are parallel.
Vector Direction
The direction of a vector is crucial in understanding how it acts in space. Simply put, the direction indicates where the vector points in a coordinate system. When we talk about the direction in mathematical terms, we often represent it by using unit vectors.
The unit vector is a vector that has a magnitude of 1 and points in the same direction as the original vector \( \mathbf{a} \), calculated by:
\[\hat{a} = \frac{\mathbf{a}}{|\mathbf{a}|}\]In our exercise, for \( \mathbf{u} \times \mathbf{v} \) which resulted in the vector \( 4\mathbf{k} \), the direction is aligned with the positive \( \mathbf{k} \)-axis. This signifies that all the vector's influence is acting in a singular, positive direction along this axis.
Conversely, \( \mathbf{v} \times \mathbf{u} \) resulting in \( -4\mathbf{k} \) tells us the direction is along the negative \( \mathbf{k} \)-axis. This negative sign simply reflects the direction opposite to the positive \( \mathbf{k} \) that was present with the first cross product result. Understanding vector direction helps visualize and predict interactions in fields like physics and engineering.
The unit vector is a vector that has a magnitude of 1 and points in the same direction as the original vector \( \mathbf{a} \), calculated by:
\[\hat{a} = \frac{\mathbf{a}}{|\mathbf{a}|}\]In our exercise, for \( \mathbf{u} \times \mathbf{v} \) which resulted in the vector \( 4\mathbf{k} \), the direction is aligned with the positive \( \mathbf{k} \)-axis. This signifies that all the vector's influence is acting in a singular, positive direction along this axis.
Conversely, \( \mathbf{v} \times \mathbf{u} \) resulting in \( -4\mathbf{k} \) tells us the direction is along the negative \( \mathbf{k} \)-axis. This negative sign simply reflects the direction opposite to the positive \( \mathbf{k} \) that was present with the first cross product result. Understanding vector direction helps visualize and predict interactions in fields like physics and engineering.
Determinant Method
The determinant method is a key technique when finding the cross product of vectors. It involves using a 3x3 matrix made up of unit vectors and the components of the vectors we are working with. The matrix is structured as follows for vectors \( \mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k} \) and \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \):
\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix}\]This matrix is expanded out using the rule of Sarrus or cofactor expansion, resulting in a vector that is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \).
\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix}\]This matrix is expanded out using the rule of Sarrus or cofactor expansion, resulting in a vector that is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \).
- The three resulting components derive from the cyclic permutation of \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).
- It can demonstrate the right-hand rule, where the direction of the product follows this heuristic rule.