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Chapter 17: Problem 36

Given \(\mathbf{A}=\langle 3,2\rangle ; \mathbf{C}=\langle 8,8\rangle ; \mathbf{A}+\mathbf{B}=\mathbf{C} ;\) find \(|\mathbf{B}|\).

Short Answer

Expert verified
\(|\textbf{B}|= \sqrt{61}\)

Step by step solution

01

Understand the Given Information

You are given vectors \(\textbf{A} = \langle 3, 2\rangle\) and \(\textbf{C} = \langle 8, 8\rangle\).\ You also have the equation \(\textbf{A} + \textbf{B} = \textbf{C}\).\ The goal is to find \(|\textbf{B}|\) .
02

Express Vector B in Terms of A and C

Using the equation \(\textbf{A} + \textbf{B} = \textbf{C}\),\renge to isolate \(\textbf{B}\): \(\textbf{B} = \textbf{C} - \textbf{A}\)
03

Compute Vector B

\(\textbf{B} = \langle 8, 8\rangle - \langle 3, 2\rangle = \langle 8 - 3, 8 - 2\rangle = \langle 5, 6\rangle\)
04

Find the Magnitude of Vector B

The magnitude of a vector \(\textbf{B} = \langle 5, 6\rangle\) is calculated as:\ \(|\textbf{B}| = \sqrt{5^2 + 6^2}\).Calculate the expression: \(\textbf{ B} = \sqrt{25 + 36} = \sqrt{61} \)

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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 a fundamental operation in vector mathematics. When adding two vectors, you combine their respective components to form a new vector. For example, if you have vectors \(\textbf{A} = \langle a_1, a_2 \rangle\) and \(\textbf{B} = \langle b_1, b_2 \rangle\), the sum of these vectors is \(\textbf{A} + \textbf{B} = \langle a_1 + b_1, a_2 + b_2 \rangle\).
  • This principle helps you understand how individual components of vectors combine.
  • It is useful in various fields such as physics, engineering, and computer science.
In the given problem, we know that \(\textbf{A} + \textbf{B} = \textbf{C}\). When we isolate \(\textbf{B}\) by subtracting \(\textbf{A}\) from \(\textbf{C}\), we use vector addition properties to perform this operation component-wise.
vector components
Each vector is defined by its components, typically in the form \(\textbf{V} = \langle x, y \rangle\).
  • The components represent the projections of the vector along the coordinate axes.
  • Breaking down vectors into components simplifies operations like addition, subtraction, and finding magnitudes.
In the exercise, vectors \(\textbf{A} = \langle 3, 2 \rangle\) and \(\textbf{C} = \langle 8, 8 \rangle\) are given in component form.
To find vector \(\textbf{B}\), we computed the difference component-wise: \(\textbf{B} = \textbf{C} - \textbf{A} = \langle 8 - 3, 8 - 2 \rangle = \langle 5, 6 \rangle\).
pythagorean theorem
The Pythagorean theorem is a crucial concept for calculating the magnitude of vectors. For a vector \(\textbf{V} = \langle x, y \rangle\), its magnitude is calculated as \(|\textbf{V}| = \sqrt{x^2 + y^2}\).
  • This formula comes directly from the Pythagorean theorem, applied to the right triangle formed by the vector's components.
  • The magnitude gives us the length of the vector, which is essential in many applications.
Using the components \(\textbf{B} = \langle 5, 6 \rangle\), we apply the Pythagorean theorem to find its magnitude: \(|\textbf{B}| = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}\).
This result shows how the components of vector \(\textbf{B}\) determine its overall length.

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Most popular questions from this chapter

Prove by vector analysis that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side.

Prove that two nonzero vectors are parallel if and only if the radian measure of the angle between them is 0 or \(\pi\).

In Exercises 13 through 18 , find the radius of curvature at any point on the given curve. $$ y=\sin ^{-1} x $$

Show that the curvature of the catenary \(y=a \cosh (x / a)\) at any point \((x, y)\) on the curve is \(a / y^{2} .\) Draw the circle of curvature at \((0, a)\). Show that the curvature \(K\) is an absolute maximum at the point \((0, a)\) without referring to \(K^{\prime}(x)\).

Find the radius of curvature at any point on the given curve. The cycloid \(x=a(t-\sin t), y=a(1-\cos t)\)
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