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Chapter 17: Problem 3

Find \(\mathbf{A} \cdot \mathbf{B}\). $$ \mathbf{A}=2 \mathbf{i}-\mathbf{j} ; \mathbf{B}=\mathbf{i}+3 \mathbf{j} $$

Short Answer

Expert verified
From given \(\textbf{A}\) and \(\textbf{B}\): \(\textbf{A} \boldsymbol{\bullet} \textbf{B} = -1\). .array.

Step by step solution

01

Express the vectors in component form

Given that \(\textbf{A} = 2 \textbf{i} - \textbf{j}\) and \(\textbf{B} = \textbf{i} + 3 \textbf{j}\), express them in component form. We write this as \(\textbf{A} = (2, -1)\) and \(\textbf{B} = (1, 3)\).
02

Use the dot product formula

The dot product formula for two vectors \( \textbf{A} = (a_1, a_2)\) and \(\textbf{B} = (b_1, b_2)\) is: \(\textbf{A} \boldsymbol{\bullet} \textbf{B} = a_1 b_1 + a_2 b_2\). In our case, \(a_1 = 2\), \(a_2 = -1\), \(b_1 = 1\), and \(b_2 = 3\).
03

Substitute the values into the formula

Substitute the values into the dot product formula: \(\textbf{A} \boldsymbol{\bullet} \textbf{B} = 2 \times 1 + (-1) \times 3\).
04

Simplify the expression

Calculate the expression: \(2 \times 1 = 2\) and \((-1) \times 3 = -3\). Therefore, \(\textbf{A} \boldsymbol{\bullet} \textbf{B} = 2 - 3\).
05

Get the final result

Simplify the result: \(2 - 3 = -1\). Hence, \(\textbf{A} \boldsymbol{\bullet} \textbf{B} = -1\).

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Key Concepts

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

vector multiplication
Vector multiplication can involve different operations like the dot product or the cross product.
For this exercise, we focus on the dot product.
It's a way of multiplying two vectors that results in a scalar value (a single number).
The dot product helps measure how much one vector extends in the direction of another.
This can be useful in physics and engineering for calculating projections and works.
dot product formula
The dot product of two vectors is calculated using their components.
For vectors \(\textbf{A} = (a_1, a_2)\) and \( \textbf{B} = (b_1, b_2)\):
\[ \textbf{A} \boldsymbol{\bullet} \textbf{B} = a_1 \times b_1 + a_2 \times b_2 \]
In the given exercise, we had: \( \textbf{A} = (2, -1) \) and \( \textbf{B} = (1, 3) \).
Using the dot product formula:
\[ \textbf{A} \boldsymbol{\bullet} \textbf{B} = 2 \times 1 + (-1) \times 3 = 2 - 3 = -1 \]
component form of vectors
Vectors are often expressed in component form to simplify calculations.
For example, for \( \textbf{A} = 2 \textbf{i} - \textbf{j} \), we write it as \( \textbf{A} = (2, -1) \).
This format breaks down the vector into its respective horizontal and vertical components.
Here's how you convert vectors into component form:
  • Identify the coefficients of \( \textbf{i} \) (horizontal) and \( \textbf{j} \) (vertical).
  • Combine these coefficients into an ordered pair.

This makes it easier to apply formulas like the dot product.

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Most popular questions from this chapter

Find parametric equations of the curve having arc length \(s\) as a parameter, where \(s\) is measured from the point where \(t=0 .\) Check your result by using Eq. (10). \(x=2+\cos t, y=3+\sin t\)

Find the radius of curvature at any point on the given curve. The tractrix \(x=t-a \tanh \frac{t}{a}, y=a \operatorname{sech} \frac{t}{a}\)

Given \(\mathbf{A}=5 \mathbf{i}+12 \mathbf{j} ; \mathbf{B}=\mathbf{i}+k \mathbf{j}\), where \(k\) is a scalar. Find \(k\) so that the radian measure of the angle between \(\mathbf{A}\) and \(\mathbf{B}\) is \(\frac{1}{3} \pi\)

A particle is moving along the curve having the given vector equation. In each problem, find the vectors \(\mathbf{V}(t), \mathbf{A}(t), \mathbf{T}(t)\), and \(\mathbf{N}(t)\), and the following scalars for an arbitrary value of \(t:|\mathbf{V}(t)|, A_{T}, A_{N}, K(t) .\) Also find the particular values when \(t=t_{1} .\) At \(t=t_{1}\), draw a sketch of a portion of the curve and representations of the vectors \(\mathbf{V}\left(t_{1}\right)\), \(\mathbf{A}\left(t_{1}\right), A_{T} \mathbf{T}\left(t_{1}\right)\), and \(A_{N} \mathbf{N}\left(t_{1}\right)\). $$ \mathbf{R}(t)=3 t^{2} \mathbf{i}+2 t^{3} \mathbf{j} ; t_{1}=1 $$

In Exercises 7 through 12, the position of a moving particle at \(t\) sec is determined from a vector equation. Find: (a) \(\mathbf{V}\left(t_{1}\right)\) (b) \(\mathbf{A}\left(t_{1}\right) ;\) (c) \(\left|\mathbf{V}\left(t_{1}\right)\right| ;\) (d) \(\left|\mathbf{A}\left(t_{1}\right)\right| .\) Draw a sketch of a portion of the path of the particle containing the position of the particle at \(t=t_{1}\), and draw the representations of \(\mathbf{V}\left(t_{1}\right)\) and \(\mathbf{A}\left(t_{1}\right)\) having initial point where \(t=t_{1}\). $$ \mathbf{R}(t)=(2 t-1) \mathbf{i}+\left(t^{2}+1\right) \mathbf{j} ; t_{1}=3 $$
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