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Chapter 7: Problem 33

In Problems \(33-36, \mathbf{a}=\langle 1,-1,3\rangle\) and \(\mathbf{b}=\langle 2,6,3\rangle .\) Find the indicated number. \(\operatorname{comp}_{\mathrm{b}}^{\mathbf{a}}\)

Short Answer

Expert verified
The component of \( \mathbf{a} \) along \( \mathbf{b} \) is \( \frac{5}{7} \).

Step by step solution

01

Understand the Problem

The problem asks for the component of vector \( \mathbf{a} \) along vector \( \mathbf{b} \). This is known as the vector projection calculation, and involves finding a scalar component.
02

Use the formula for the scalar component of a vector onto another

The formula to find the component of a vector \( \mathbf{a} \) along \( \mathbf{b} \) is given by: \[ \operatorname{comp}_{\mathbf{b}}^{\mathbf{a}} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|} \]Here, \( \mathbf{a} \cdot \mathbf{b} \) is the dot product, and \( \|\mathbf{b}\| \) is the magnitude of vector \( \mathbf{b} \).
03

Calculate the Dot Product \( \mathbf{a} \cdot \mathbf{b} \)

The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = (1)(2) + (-1)(6) + (3)(3) \]Simplifying gives:\[ = 2 - 6 + 9 = 5 \]
04

Calculate the Magnitude of Vector \( \mathbf{b} \)

The magnitude of vector \( \mathbf{b} \) is found using the formula:\[ \| \mathbf{b} \| = \sqrt{2^2 + 6^2 + 3^2} \]This simplifies to:\[ \| \mathbf{b} \| = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]
05

Compute the Scalar Component \( \operatorname{comp}_{\mathbf{b}}^{\mathbf{a}} \)

Using the values calculated:\[ \operatorname{comp}_{\mathbf{b}}^{\mathbf{a}} = \frac{5}{7} \]

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Key Concepts

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

Understanding the Dot Product
The dot product is a fundamental operation in vector math. It involves multiplying two vectors to obtain a scalar. To calculate the dot product, you multiply corresponding components from two vectors and then add them together. Given vectors \( \mathbf{a} = \langle 1, -1, 3 \rangle \) and \( \mathbf{b} = \langle 2, 6, 3 \rangle \), the formula for the dot product is:
  • \( \mathbf{a} \cdot \mathbf{b} = (1)(2) + (-1)(6) + (3)(3) \)
  • Perform each multiplication: \( 1 \times 2 = 2 \), \( -1 \times 6 = -6 \), and \( 3 \times 3 = 9 \).
  • Add these products together to get the final result: \( 2 - 6 + 9 = 5 \).
This scalar result, 5, tells us how much of \( \mathbf{a} \) is in the direction of \( \mathbf{b} \). The dot product is pivotal for further calculations such as finding projections or angles between vectors.
Magnitude of a Vector
The magnitude of a vector is essentially its length in space. For a vector \( \mathbf{b} = \langle 2, 6, 3 \rangle \), the magnitude can be calculated using the formula:
  • \( \| \mathbf{b} \| = \sqrt{x^2 + y^2 + z^2} \).
  • Plug in the components of \( \mathbf{b} \): \( x = 2, y = 6, z = 3 \).
  • Calculate each squared component: \( 2^2 = 4 \), \( 6^2 = 36 \), \( 3^2 = 9 \).
  • Sum these squares: \( 4 + 36 + 9 = 49 \).
  • Take the square root to find the magnitude: \( \sqrt{49} = 7 \).
The magnitude, 7, gives a measure of vector \( \mathbf{b} \)'s length and is needed to compute the scalar component of one vector along another.
Scalar Component of a Vector
The scalar component, or projection of one vector onto another, tells us how much of one vector lies in the direction of another vector. To find this, we employ:
  • Formula: \( \operatorname{comp}_{\mathbf{b}}^{\mathbf{a}} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|} \).
  • The dot product \( \mathbf{a} \cdot \mathbf{b} \) was calculated as 5.
  • The magnitude \( \|\mathbf{b}\| \) was found to be 7.
  • Divide the dot product by the magnitude: \( \frac{5}{7} \).
This gives a scalar value of \( \frac{5}{7} \), representing how much of \( \mathbf{a} \) aligns with \( \mathbf{b} \) in terms of length. This concept is especially useful in physics and engineering for analyzing forces and movements.

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Most popular questions from this chapter

Use the dot product to prove the triangle inequality \(\|\mathbf{a}+\mathbf{b}\| \leq\|\mathbf{a}\|+\|\mathbf{b}\| .\) [Hint: Consider property \((v i)\) of the dot product.]

The vectors \(\mathbf{u}_{1}=\langle 1,0,0\rangle, \mathbf{u}_{2}=\langle 1,1,0\rangle\), and \(\mathbf{u}_{3}=\langle 1,1,1\rangle\) form a basis for the vector space \(R^{3}\). (a) Show that \(\mathbf{u}_{1}, \mathbf{u}_{2}\), and \(\mathbf{u}_{3}\) are linearly independent. (b) Express the vector \(\mathbf{a}=\langle 3,-4,8\rangle\) as a linear combination of \(\mathbf{u}_{1}, \mathbf{u}_{2}\), and \(\mathbf{u}_{3}\).

Let \(\mathscr{L}_{a}\) and \(\mathscr{L}_{b}\) be lines with direction vectors a and \(\mathbf{b}\), respectively. \(\mathscr{L}_{a}\) and \(\mathscr{L}_{b}\) areorthogonalifa and \(\mathbf{b}\) are orthogonal and parallel if a and \(\mathbf{b}\) are parallel. Determine which of the following lines are orthogonal and which are parallel. (a) \(\mathbf{r}=\langle 1,0,2\rangle+t\langle 9,-12,6\rangle\) (b) \(x=1+9 t, y=12 t, z=2-6 t\) (c) \(x=2 t, y=-3 t, z=4 t\) (d) \(x=5+t, y=4 t, z=3+\frac{5}{2} t\) (e) \(x=1+t, y=\frac{3}{2} t, z=2-\frac{3}{2} t\) (f) \(\frac{x+1}{-3}=\frac{y+6}{4}=\frac{z-3}{-2}\)

In Problems 37 and 38, find the component of the given vector in the direction from the origin to the indicated point. \(\mathbf{a}=4 \mathbf{i}+6 \mathbf{j}, \quad P(3,10)\)

Find a unit vector in the same direction as \(\mathbf{a}=\mathbf{i}-3 \mathbf{j}+2 \mathbf{k}\).
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