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Chapter 14: Problem 4

Use the definition of vector products to verify \(\mathbf{u} \times(\mathbf{v} \pm \mathbf{w})=(\mathbf{u} \times \mathbf{v}) \pm(\mathbf{u} \times \mathbf{w}) .\) (This is the distributive property of vector product over addition and subtraction.)

Short Answer

Expert verified
The cross product is distributive over addition and subtraction: \(\mathbf{u} \times (\mathbf{v} \pm \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) \pm (\mathbf{u} \times \mathbf{w})\).

Step by step solution

01

Define the Vector Cross Product

The vector cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by the determinant formula: \[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3\end{vmatrix} = ( a_2b_3 - a_3b_2 ) \mathbf{i} - ( a_1b_3 - a_3b_1 ) \mathbf{j} + ( a_1b_2 - a_2b_1 ) \mathbf{k}\]
02

Substitute into the Left-Hand Side

Substitute \( \mathbf{v} \pm \mathbf{w} = \langle v_1 \pm w_1, v_2 \pm w_2, v_3 \pm w_3 \rangle \) into the cross product \( \mathbf{u} \times (\mathbf{v} \pm \mathbf{w}) \), applying the determinant formula: \[\mathbf{u} \times (\mathbf{v} \pm \mathbf{w}) = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \u_1 & u_2 & u_3 \v_1 \pm w_1 & v_2 \pm w_2 & v_3 \pm w_3\end{vmatrix}\] This becomes \[=\left( u_2(v_3 \pm w_3) - u_3(v_2 \pm w_2) \right) \mathbf{i} - \left( u_1(v_3 \pm w_3) - u_3(v_1 \pm w_1) \right) \mathbf{j} + \left( u_1(v_2 \pm w_2) - u_2(v_1 \pm w_1) \right) \mathbf{k}\]
03

Expand the Expression

Expand the expression from Step 2:\[=\left( u_2v_3 \pm u_2w_3 - u_3v_2 \mp u_3w_2 \right) \mathbf{i} - \left( u_1v_3 \pm u_1w_3 - u_3v_1 \mp u_3w_1 \right) \mathbf{j} + \left( u_1v_2 \pm u_1w_2 - u_2v_1 \mp u_2w_1 \right) \mathbf{k}\]
04

Compute the Right-Hand Side

Compute \((\mathbf{u} \times \mathbf{v}) \pm (\mathbf{u} \times \mathbf{w})\) using the cross product formula for each pair separately and then add/subtract:For \( \mathbf{u} \times \mathbf{v} \): \[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \u_1 & u_2 & u_3 \v_1 & v_2 & v_3\end{vmatrix} = (u_2v_3 - u_3v_2) \mathbf{i} - (u_1v_3 - u_3v_1) \mathbf{j} + (u_1v_2 - u_2v_1) \mathbf{k}\]For \( \mathbf{u} \times \mathbf{w} \): \[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \u_1 & u_2 & u_3 \w_1 & w_2 & w_3\end{vmatrix} = (u_2w_3 - u_3w_2) \mathbf{i} - (u_1w_3 - u_3w_1) \mathbf{j} + (u_1w_2 - u_2w_1) \mathbf{k}\]Add/Subtract the corresponding components.
05

Compare Both Sides

Compare the expanded expression from the Left-Hand Side with the Right-Hand Side results:Left-Hand Side from Step 3:\[\left( u_2v_3 \pm u_2w_3 - u_3v_2 \mp u_3w_2 \right) \mathbf{i} - \left( u_1v_3 \pm u_1w_3 - u_3v_1 \mp u_3w_1 \right) \mathbf{j} + \left( u_1v_2 \pm u_1w_2 - u_2v_1 \mp u_2w_1 \right) \mathbf{k}\]Right-Hand Side from Step 4:\[\left((u_2v_3 - u_3v_2) \pm (u_2w_3 - u_3w_2) \right) \mathbf{i} - \left((u_1v_3 - u_3v_1) \pm (u_1w_3 - u_3w_1) \right) \mathbf{j} + \left((u_1v_2 - u_2v_1) \pm (u_1w_2 - u_2w_1) \right) \mathbf{k}\]Both expressions are identical, confirming the distributive property.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distributive Property of Vector Products
The distributive property in vector mathematics is much like that in basic arithmetic, but with vectors, it involves the cross product. When we say that the vector cross product is distributive over addition and subtraction, it means you can expand expressions like
  • \( \mathbf{u} \times (\mathbf{v} + \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) + (\mathbf{u} \times \mathbf{w}) \)
  • \( \mathbf{u} \times (\mathbf{v} - \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) - (\mathbf{u} \times \mathbf{w}) \)
This property allows you to break down complex vector products into simpler parts, making calculations easier to handle. To verify this property, you can use the determinant formula for cross products to calculate manually, comparing the expanded forms of both sides of your equation.
Foundations of Vector Mathematics
Vector mathematics is vital for various fields such as physics, engineering, and computer graphics. A vector is represented in three dimensions and has both a direction and magnitude. Vector operations follow certain algebraic rules and have unique properties compared to scalar operations. Three primary vector operations are:
  • Addition: Simply add corresponding components of the vectors.
  • Scalar multiplication: Multiply each component by the scalar (a number).
  • Cross product: This operation involves determinants and is used to find a vector perpendicular to the plane containing two original vectors.
The cross product, in particular, is useful for calculations involving torque and angular momentum, making an understanding of it essential for practical applications.
Properties of the Cross Product
The cross product is a bit more complex than other vector operations due to its unique properties. Several key properties include:
  • Distributive Property: As shown, the cross product distributes over addition.
  • Anticommutative Property: The order matters, \( \mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}) \).
  • Zero Vector Result: If two vectors are parallel, their cross product is zero, \( \mathbf{a} \times \mathbf{a} = \mathbf{0} \).
  • Magnitude: The magnitude of \( \mathbf{a} \times \mathbf{b} \) is equal to \( |\mathbf{a}||\mathbf{b}| \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).
These properties are powerful tools for working through problems that involve vector mathematics, offering both conceptual understanding and practical calculation techniques.

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Most popular questions from this chapter

Let \(\mathbf{u}=\mathbf{i}+3 \mathbf{j}-2 \mathbf{k}\) and \(\mathbf{v}=2 \mathbf{i}+\mathbf{j} .\) Find a) $$|\mathbf{u}+\mathbf{v}|$$ b) $$|\mathbf{u}|+|\mathbf{v}|$$ c) $$|-3 u|+|3 v|$$ d) $$\frac{1}{|\mathbf{u}|} \mathbf{u}$$ e) \(\left|\frac{1}{|\mathbf{u}|} \mathbf{u}\right|\)

Show that \(|\mathbf{u} \times \mathbf{v}|=\sqrt{|\mathbf{u}|^{2}|\mathbf{v}|^{2}-(\mathbf{u} \cdot \mathbf{v})^{2}}\).

Consider the following equations representing the paths of cars after starting time \(t \geqslant 0,\) where distances are measured in \(\mathrm{km}\) and time in hours. For each car, determine (i) starting position (ii) the velocity vector (iii) the speed. a) \(r=(3,-4)+t\left(\begin{array}{r}7 \\ 24\end{array}\right)\). b) \(\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{r}-3 \\\ 1\end{array}\right)+t\left(\begin{array}{r}5 \\ -12\end{array}\right)\). c) \((x, y)=(5,-2)+t(24,-7)\).

Shortly after take-oft, a plane is rising at a rate of \(300 \mathrm{m} / \mathrm{min}\). It is heading at an angle of \(45^{\circ}\) north-west with an airspeed of \(200 \mathrm{km} / \mathrm{h}\). Find the components of its velocity vector. The \(x\) -axis is in the east direction, the \(y\) -axis north and the z-axis is the elevation.

Find vectors that satisfy the stated conditions: a) opposite direction of \(\mathbf{u}=(-3,4)\) and third the magnitude of \(\mathbf{u}\) b) length of 12 and same direction as \(\mathbf{w}=4 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}\) c) of the form \(x \mathbf{i}+y \mathbf{j}-2 \mathbf{k}\) and parallel to \(\mathbf{w}=\mathbf{i}-4 \mathbf{j}+3 \mathbf{k}\)
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