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Two vectors are given as \(\mathbf{b}=(1,2,3)\) and \(\mathbf{c}=(3,2,1)\). (Remember that these statements are just a compact way of giving you the components of the vectors.) Find \(\mathbf{b}+\mathbf{c}, 5 \mathbf{b}-2 \mathbf{c}, \mathbf{b} \cdot \mathbf{c},\) and \(\mathbf{b} \times \mathbf{c}\).

Short Answer

Expert verified
\(\mathbf{b} + \mathbf{c} = (4, 4, 4)\); \(5\mathbf{b} - 2\mathbf{c} = (-1, 6, 13)\); \(\mathbf{b} \cdot \mathbf{c} = 10\); \(\mathbf{b} \times \mathbf{c} = (-4, 8, -4)\).

Step by step solution

01

Add the vectors \(\mathbf{b}\) and \(\mathbf{c}\)

To find the sum \(\mathbf{b} + \mathbf{c}\), simply add the corresponding components of the vectors. \[ \mathbf{b} + \mathbf{c} = (1+3, 2+2, 3+1) = (4, 4, 4) \]
02

Subtract scaled vectors \(5\mathbf{b} - 2\mathbf{c}\)

First, scale each vector by the given coefficients: \(5\mathbf{b} = 5(1, 2, 3) = (5, 10, 15)\)\(2\mathbf{c} = 2(3, 2, 1) = (6, 4, 2)\)Then subtract the components of \(2\mathbf{c}\) from \(5\mathbf{b}\):\[ 5\mathbf{b} - 2\mathbf{c} = (5-6, 10-4, 15-2) = (-1, 6, 13) \]
03

Compute dot product \(\mathbf{b} \cdot \mathbf{c}\)

The dot product is obtained by multiplying corresponding components and summing the results:\[ \mathbf{b} \cdot \mathbf{c} = (1 \cdot 3) + (2 \cdot 2) + (3 \cdot 1) = 3 + 4 + 3 = 10 \]
04

Compute cross product \(\mathbf{b} \times \mathbf{c}\)

The cross product of two vectors in three dimensions is given by\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 3 \ 3 & 2 & 1 \end{vmatrix} \]Calculate the determinant:- For \(\mathbf{i}\)-component: \((2)(1) - (3)(2) = 2 - 6 = -4\)- For \(\mathbf{j}\)-component: \((1)(1) - (3)(3) = 1 - 9 = -8\), note the sign change makes it \(+8\)- For \(\mathbf{k}\)-component: \((1)(2) - (2)(3) = 2 - 6 = -4\)Thus, \[ \mathbf{b} \times \mathbf{c} = (-4, 8, -4) \]

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Key Concepts

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

Vector Addition
Vector addition is one of the fundamental operations in vector algebra. It is performed component-wise. Let's take two vectors, \( \mathbf{b} = (1, 2, 3) \) and \( \mathbf{c} = (3, 2, 1) \). To find their sum, \( \mathbf{b} + \mathbf{c} \), you simply add their corresponding components together. This means adding the x-components, y-components, and z-components separately.

The resulting vector \( \mathbf{b} + \mathbf{c} \) is:

\[ \mathbf{b} + \mathbf{c} = (1 + 3, 2 + 2, 3 + 1) = (4, 4, 4) \]

Vector addition can be visualized as placing the second vector at the tip of the first vector and then drawing the resultant vector from the origin to the tip of the second vector. This is often referred to as the "head-to-tail" method.

Key points:
  • Always add corresponding components.
  • The result is another vector.
  • Visualize it geometrically as connecting vectors from head to tail.
Dot Product
The dot product, also known as the scalar product, is an operation that combines two vectors to produce a scalar quantity. For vectors \( \mathbf{b} = (1, 2, 3) \) and \( \mathbf{c} = (3, 2, 1) \), the dot product is calculated by multiplying each corresponding pair of components and then summing those products. This operation is essential in various applications such as physics, computer graphics, and finding angles between vectors.

Here's how you calculate it:

\[ \mathbf{b} \cdot \mathbf{c} = (1 \cdot 3) + (2 \cdot 2) + (3 \cdot 1) = 3 + 4 + 3 = 10 \]

Key aspects of the dot product:
  • Produces a scalar, not a vector.
  • Used to determine if vectors are orthogonal (dot product is zero).
  • Provides a measure of the vectors' directional alignment.
Cross Product
The cross product is a vector operation that results in another vector that is perpendicular to the plane containing the original vectors. For vectors \( \mathbf{b} = (1, 2, 3) \) and \( \mathbf{c} = (3, 2, 1) \), the cross product is calculated using a determinant. This operation is crucial in physics and engineering for finding torques, areas of parallelograms, and more.

Calculation involves determining the determinant of a matrix composed of both vectors and the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \):

\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 3 \ 3 & 2 & 1 \end{vmatrix} \]
The components are:
  • \( \mathbf{i}\)-component: \((2)(1) - (3)(2) = -4\)
  • \( \mathbf{j}\)-component: \((1)(1) - (3)(3) = -8\) with sign change to \(+8\)
  • \( \mathbf{k}\)-component: \((1)(2) - (2)(3) = -4\)
Thus, the resultant vector is:

\[ \mathbf{b} \times \mathbf{c} = (-4, 8, -4) \]
Features of the cross product:
  • Results in a vector perpendicular to the original vectors.
  • Magnitude gives the area of the parallelogram formed by the vectors.
  • Direction follows the right-hand rule.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a single real number). This operation changes the magnitude of the vector but not its direction, unless the scalar is negative, which would reverse the direction. In the given exercise, we perform this operation on vectors \( \mathbf{b} = (1, 2, 3) \) and \( \mathbf{c} = (3, 2, 1) \).

For vector \( \mathbf{b} \), multiply each component by 5:
\[ 5\mathbf{b} = 5 \times (1, 2, 3) = (5, 10, 15) \]
For vector \( \mathbf{c} \), multiply each component by 2:
\[ 2\mathbf{c} = 2 \times (3, 2, 1) = (6, 4, 2) \]
These transformed vectors can further undergo other operations, such as subtraction, resulting in:
\[ 5\mathbf{b} - 2\mathbf{c} = (5 - 6, 10 - 4, 15 - 2) = (-1, 6, 13) \]
Important points about scalar multiplication:
  • Changes magnitude, not the direction (unless negative).
  • Scaling positive enlarges, scaling negative reverses direction.
  • Scalars can be used to resolve vectors into desired lengths.

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Most popular questions from this chapter

[Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from \(\phi_{\mathrm{o}}=20\) degrees, using the values \(R=5 \mathrm{m}\) and \(g=9.8 \mathrm{m} / \mathrm{s}^{2} .\) Make a plot of \(\phi\) against time for two or three periods. ( \(\mathbf{b}\) ) On the same picture, plot the approximate solution (1.57) with the same \(\phi_{\mathrm{o}}=20^{\circ}\) Comment on your two graphs. Note: If you haven't used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for "NDSolve鈥漚nd how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.

If you have some experience in electromagnetism, you could do the following problem concerning the curious situation illustrated in Figure \(1.8 .\) The electric and magnetic fields at a point \(\mathbf{r}_{1}\) due to a charge \(q_{2}\) at \(\mathbf{r}_{2}\) moving with constant velocity \(\mathbf{v}_{2}\) (with \(v_{2} \ll c\) ) are \(^{15}\) $$\mathbf{E}\left(\mathbf{r}_{1}\right)=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{q_{2}}{s^{2}} \hat{\mathbf{s}} \quad \text { and } \quad \mathbf{B}\left(\mathbf{r}_{1}\right)=\frac{\mu_{0}}{4 \pi} \frac{q_{2}}{s^{2}} \mathbf{v}_{2} \times \hat{\mathbf{s}}$$ where \(\mathbf{s}=\mathbf{r}_{1}-\mathbf{r}_{2}\) is the vector pointing from \(\mathbf{r}_{2}\) to \(\mathbf{r}_{1}\). (The first of these you should recognize as Coulomb's law.) If \(\mathbf{F}_{12}^{\mathrm{el}}\) and \(\mathbf{F}_{12}^{\text {mag }}\) denote the electric and magnetic forces on a charge \(q_{1}\) at \(\mathbf{r}_{1}\) with velocity \(\mathbf{v}_{1},\) show that \(F_{12}^{\operatorname{mag}} \leq\left(v_{1} v_{2} / c^{2}\right) F_{12}^{\mathrm{el}} .\) This shows that in the non-relativistic domain it is legitimate to ignore the magnetic force between two moving charges.

A cannon shoots a ball at an angle \(\theta\) above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with \(x\) measured horizontally and \(y\) vertically.) (b) Let \(r(t)\) denote the ball's distance from the cannon. What is the largest possible value of \(\theta\) if \(r(t)\) is to increase throughout the ball's flight? [Hint: Using your solution to part (a) you can write down \(r^{2}\) as \(x^{2}+y^{2},\) and then find the condition that \(r^{2}\) is always increasing.]

Prove that the two definitions of the scalar product \(\mathbf{r} \cdot\) s as \(r s \cos \theta(1.6)\) and \(\sum r_{i} s_{i}(1.7)\) are equal. One way to do this is to choose your \(x\) axis along the direction of \(\mathbf{r}\). [Strictly speaking you should first make sure that the definition (1.7) is independent of the choice of axes. If you like to worry about such niceties, see Problem 1.16.]

You lay a rectangular board on the horizontal floor and then tilt the board about one edge until it slopes at angle \(\theta\) with the horizontal. Choose your origin at one of the two corners that touch the floor, the \(x\) axis pointing along the bottom edge of the board, the \(y\) axis pointing up the slope, and the \(z\) axis normal to the board. You now kick a frictionless puck that is resting at \(O\) so that it slides across the board with initial velocity \(\left(v_{\mathrm{ox}}, v_{\mathrm{oy}}, 0\right) .\) Write down Newton's second law using the given coordinates and then find how long the puck takes to return to the floor level and how far it is from \(O\) when it does so.

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