Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 12: Problem 36
\(35-40\) Find the scalar and vector projections of b onto a. $$\mathbf{a}=\langle 1,2\rangle, \quad \mathbf{b}=\langle- 4,1\rangle$$
Short Answer
Expert verified
Scalar projection is \(-\frac{2}{\sqrt{5}}\), vector projection is \(\langle -\frac{2}{5}, -\frac{4}{5} \rangle\).
Step by step solution
01
Calculate the Dot Product of Vectors a and b
The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \). Substitute the given values: \( \mathbf{a} = \langle 1, 2 \rangle \) and \( \mathbf{b} = \langle -4, 1 \rangle \). Calculate: \( \mathbf{a} \cdot \mathbf{b} = (1)(-4) + (2)(1) = -4 + 2 = -2 \).
02
Calculate the Magnitude of Vector a
The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is calculated using the formula \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \). Substitute the components of \( \mathbf{a} \):\( \| \mathbf{a} \| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \).
03
Determine the Scalar Projection of b onto a
The scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by the formula \( \text{comp}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|} \).Substitute the values: \( \text{comp}_{\mathbf{a}} \mathbf{b} = \frac{-2}{\sqrt{5}} \).
04
Calculate the Vector Projection of b onto a
The vector projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by the formula \( \text{proj}_{\mathbf{a}} \mathbf{b} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|^2} \right) \mathbf{a} \).First, calculate \( \| \mathbf{a} \|^2 = (\sqrt{5})^2 = 5 \).Then, substitute the dot product and the vector \( \mathbf{a} \):\( \text{proj}_{\mathbf{a}} \mathbf{b} = \left( \frac{-2}{5} \right) \mathbf{a} = \left( \frac{-2}{5} \right) \langle 1, 2 \rangle = \langle -\frac{2}{5}, -\frac{4}{5} \rangle \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
The dot product is a way to multiply two vectors and get a scalar (a single number) as a result. It's calculated by multiplying the corresponding components of the two vectors and then adding those products together. In mathematical terms, if you have two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is given by:
- \( \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2 \)
- \( \mathbf{a} \cdot \mathbf{b} = (1)(-4) + (2)(1) = -4 + 2 = -2 \)
Magnitude of Vector
The magnitude of a vector represents its length or size, providing a measure of the vector's extent in space. To find the magnitude of a vector like \( \mathbf{a} = \langle a_1, a_2 \rangle \), you use the following formula:
- \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \)
- \( \| \mathbf{a} \| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \)
Scalar Projection
Scalar projection helps to find how much of one vector lies in the direction of another. It's like asking "how much of vector \( \mathbf{b} \) is pointing in the direction of vector \( \mathbf{a} \)?" The scalar projection, or component, is calculated using the formula:
- \( \text{comp}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|} \)
- \( \text{comp}_{\mathbf{a}} \mathbf{b} = \frac{-2}{\sqrt{5}} \)
Projection Formula
The projection formula extends the idea of scalar projection by introducing a vector, offering insight into both the magnitude and direction of the projection. Calculating the vector projection \( \text{proj}_{\mathbf{a}} \mathbf{b} \) involves the following formula:
- \( \text{proj}_{\mathbf{a}} \mathbf{b} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|^2} \right) \mathbf{a} \)
- \( \| \mathbf{a} \|^2 = (\sqrt{5})^2 = 5 \)
- \( \text{proj}_{\mathbf{a}} \mathbf{b} = \left( \frac{-2}{5} \right) \langle 1, 2 \rangle = \langle -\frac{2}{5}, -\frac{4}{5} \rangle \)
