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$$[(2 \mathbf{k}) \times(3 \mathbf{j})] \times(4 \mathbf{j})=[2 \cdot 3(\mathbf{k} \times \mathbf{j}) \times(4 \mathbf{j})]=6(-\mathbf{i}) \times 4 \mathbf{j}=(-6)(4)(\mathbf{i} \times \mathbf{j})=-24 \mathbf{k}$$

Short Answer

Expert verified
The solution is \(-24 \mathbf{k}\).

Step by step solution

01

Calculate the first cross product

The vector \(2 \mathbf{k}\) is crossed with \(3 \mathbf{j}\). In the standard unit vector notation, the cross product of two vectors \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) and \(\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}\) is determined by the determinant: \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\).For \(2 \mathbf{k}\) and \(3 \mathbf{j}\), we have \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 0 & 2 \ 0 & 3 & 0 \end{vmatrix} = 6(-\mathbf{i})\).
02

Calculate the second cross product

Now take the result from Step 1, \(6(-\mathbf{i})\), and cross it with \(4 \mathbf{j}\). The cross product gives: \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -6 & 0 & 0 \ 0 & 4 & 0 \end{vmatrix} = (-6)(0 - 0) \mathbf{i} - (0)(0 - (-6 \cdot 0)) \mathbf{j} + ((-6)(4) - 0) \mathbf{k}\] Simplifying, we get \(-24 \mathbf{k}\).
03

Verify the components

Verify calculations by checking each multiplier in the cross product calculations from Steps 1 and 2. Cross-check signs and magnitudes in each step:1. Cross product \(6(\mathbf{j} \times \mathbf{k}) = 6(-\mathbf{i})\) is correct from Step 1.2. Cross product \((-6\mathbf{i}) \times 4 \mathbf{j} = -24 \mathbf{k}\) in Step 2 confirms the results.Recheck determinant to ensure there is consistency with vector identities.

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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 way to multiply two vectors in three-dimensional space. Its result is a vector that is perpendicular to both of the original vectors. This operation is particularly useful in physics and engineering, especially when dealing with rotations or torque.

To compute the cross product of vectors \( \mathbf{a} \) and \( \mathbf{b} \), we can use the determinant of a specific 3x3 matrix. With vectors expressed in terms of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), we place these unit vectors in the first row of the matrix. The components of \( \mathbf{a} \) and \( \mathbf{b} \) follow in the subsequent rows. This looks like:

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \]

The columns thus correspond to the vector's components in each direction.

From this, we can derive specific unit vector identities, such as:
  • \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \)
  • \( \mathbf{j} \times \mathbf{k} = \mathbf{i} \)
  • \( \mathbf{k} \times \mathbf{i} = \mathbf{j} \)
These identities show the cyclic nature of the cross product when using unit vectors.
Unit Vectors
Unit vectors are vectors with a magnitude of exactly one unit. They are essential in vector calculus because they provide direction without being influenced by magnitude. In three-dimensional space, the primary unit vectors are:\( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), which point along the x, y, and z axes, respectively.

Every vector in space can be expressed as a combination of these three unit vectors:

\[ \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \]

where \( a, b, \) and \( c \) are the magnitudes in each direction. This representation is crucial because it simplifies many calculations involving vectors, such as addition, scaling, and especially the determination of cross and dot products.

In the context of the problem solved, unit vectors help articulate precisely how vectors interact in space. For instance, when forming a cross product, unit vectors reveal how one vector might rotate or scale concerning another.
Determinants
The determinant is a numerical value computed from a square matrix. In the context of cross products, determinants help compute the resulting vector by encapsulating the linear combinations of the matrix's components. This is essential because it captures all interactions between vector components in a concise form.

The determinant of a 3x3 matrix, for example, is calculated as:

\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]for a matrix

\[ \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \]

In cross product calculations, the determinant enables the isolation of each component to find the resultant vector. Calculating these determinants accurately is vital for ensuring the correct direction and magnitude of the product.

Understanding determinants also aids in verifying vector transformations and properties, such as confirming that the cross product result is orthogonal to the original vectors.

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Most popular questions from this chapter

$$\|\mathbf{a}\|=\sqrt{1+9+4}=\sqrt{14} ; \quad \mathbf{u}=\frac{1}{\sqrt{14}}(\mathbf{i}-3 \mathbf{j}+2 \mathbf{k})=\frac{1}{\sqrt{14}} \mathbf{i}-\frac{3}{\sqrt{14}} \mathbf{j}+\frac{2}{\sqrt{14}} \mathbf{k}$$

From \(y^{\prime}=\frac{1}{2} x\) we see that the slope of the tangent line at (2,2) is \(1 .\) A vector with slope 1 is \(\mathbf{i}+\mathbf{j} .\) A unit vector is \((\mathbf{i}+\mathbf{j}) /\|\mathbf{i}+\mathbf{j}\|=(\mathbf{i}+\mathbf{j}) / \sqrt{2}=\frac{1}{\sqrt{2}} \mathbf{i}+\frac{1}{\sqrt{2}} \mathbf{j} .\) Another unit vector tangent to the curve is \(-\frac{1}{\sqrt{2}} \mathbf{i}-\frac{1}{\sqrt{2}} \mathbf{j}\)

$$\begin{array}{l} \mathbf{a} \times(\mathbf{b}+\mathbf{c})=\left|\begin{array}{cc} a_{2} & a_{3} \\ b_{2}+c_{2} & b_{3}+c_{3} \end{array}\right| \mathbf{i}-\left|\begin{array}{cc} a_{1} & a_{3} \\ b_{1}+c_{1} & b_{3}+c_{3} \end{array}\right| \mathbf{j}+\left|\begin{array}{cc} a_{1} & a_{2} \\ b_{1}+c_{1} & b_{2}+c_{2} \end{array}\right| \mathbf{k} \\ =\left(a_{2} b_{3}-a_{3} b_{2}\right) \mathbf{i}+\left(a_{2} c_{3}-a_{3} c_{2}\right) \mathbf{i}-\left[\left(a_{1} b_{3}-a_{3} b_{1}\right) \mathbf{j}+\left(a_{1} c_{3}-a_{3} c_{1}\right) \mathbf{j}\right]+\left(a_{1} b_{2}-a_{2} b_{1}\right) \mathbf{k}+\left(a_{1} c_{2}-a_{2} c_{1}\right) \mathbf{k} \\ =\left(a_{2} b_{3}-a_{3} b_{2}\right) \mathbf{i}-\left(a_{1} b_{3}-a_{3} b_{1}\right) \mathbf{j}+\left(a_{1} b_{2}-a_{2} b_{1}\right) \mathbf{k}+\left(a_{2} c_{3}-a_{3} c_{2}\right) \mathbf{i}-\left(a_{1} c_{3}-a_{3} c_{1}\right) \mathbf{j}+\left(a_{1} c_{2}-a_{2} c_{1}\right) \mathbf{k} \\ =\mathbf{a} \times \mathbf{b}+\mathbf{a} \times \mathbf{c} \end{array}$$

In Problems 73 and 74, the cross product of the normal vectors to the two planes will be a vector parallel to both planes, and hence a direction vector for a line parallel to the two planes. Normal vectors are \langle 1,1,-4\rangle and \(\langle 2,-1,1\rangle .\) A direction vector is \\[\langle 1,1,-4\rangle \times\langle 2,-1,1\rangle=\langle-3,-9,-3\rangle=-3\langle 1,3,1\rangle\\] Equations of the line are \(x=5+t, y=6+3 t, z=-12+t\).

$$\begin{array}{l} \mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\mathbf{b} \times(\mathbf{c} \times \mathbf{a})+\mathbf{c} \times(\mathbf{a} \times \mathbf{b}) \\ \quad=(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}+(\mathbf{b} \cdot \mathbf{a}) \mathbf{c}-(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}+(\mathbf{c} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{c} \cdot \mathbf{a}) \mathbf{b} \\ \quad=[(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}]+[(\mathbf{b} \cdot \mathbf{a}) \mathbf{c}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}]+[(\mathbf{c} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}]=\mathbf{0} \end{array}$$

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