Chapter 12: Problem 39
\(39-44\) Find the scalar and vector projections of \(\mathbf{b}\) onto \(\mathbf{a}\) $$ \mathbf{a}=\langle- 5,12\rangle, \quad \mathbf{b}=\langle 4,6\rangle $$
Short Answer
Expert verified
Scalar projection is 4; vector projection is \( \left\langle -\frac{260}{169}, \frac{624}{169} \right\rangle \).
Step by step solution
01
Calculate the Dot Product of Vectors
The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \). For the vectors \( \mathbf{a} = \langle -5, 12 \rangle \) and \( \mathbf{b} = \langle 4, 6 \rangle \), this becomes:\[ \mathbf{a} \cdot \mathbf{b} = (-5)(4) + (12)(6) = -20 + 72 = 52 \]
02
Find the Magnitude of the Vector \( \mathbf{a} \)
The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is given by \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \). For \( \mathbf{a} = \langle -5, 12 \rangle \), compute:\[ \| \mathbf{a} \| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]
03
Compute the Scalar Projection
The scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by the formula:\[ \text{Scalar projection} = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|} \]Using the results from Steps 1 and 2, this becomes:\[ \text{Scalar projection} = \frac{52}{13} = 4 \]
04
Compute the Vector Projection
The vector projection of \( \mathbf{b} \) onto \( \mathbf{a} \) uses the formula:\[ \text{Vector projection} = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|^2} \mathbf{a} \]First, compute \( \| \mathbf{a} \|^2 \):\[ \| \mathbf{a} \|^2 = 169 \]Then, the vector projection is:\[ \text{Vector projection} = \frac{52}{169} \times \langle -5, 12 \rangle = \left\langle \frac{52 \times (-5)}{169}, \frac{52 \times 12}{169} \right\rangle = \left\langle -\frac{260}{169}, \frac{624}{169} \right\rangle \]
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
The dot product is a fundamental operation in vector mathematics. It's used to determine the relationship between two vectors. Given vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is calculated as: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \] This results in a scalar, which can reveal important geometric properties, such as:
- The angle between the vectors (a positive dot product indicates an acute angle, a negative indicates obtuse)
- Whether the vectors are perpendicular (when the dot product is zero)
Magnitude of a Vector
The magnitude of a vector measures its length or size. For a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), its magnitude is calculated using the formula: \[ \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \] This formula comes from the Pythagorean theorem, applied in the context of vector components.
- The magnitude is always a non-negative quantity.
- It reflects the Euclidean distance from the origin to the point \( \mathbf{a} \).
Scalar Projection
The scalar projection of one vector onto another is like casting a shadow. It represents the length of this shadow on the direction of the other vector. The scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by: \[ \text{Scalar projection} = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|} \] This formula balances the dot product with the magnitude of the vector \( \mathbf{a} \), producing a scalar that describes the component of \( \mathbf{b} \) in the direction of \( \mathbf{a} \).
- A positive result indicates a similar direction.
- A negative indicates an opposite direction.
Vector Projection
The vector projection of \( \mathbf{b} \) onto \( \mathbf{a} \) provides a vector that reflects how much of \( \mathbf{b} \) is in the direction of \( \mathbf{a} \). Calculated as: \[ \text{Vector projection} = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|^2} \mathbf{a} \]This formula multiplies the unit vector in the direction of \( \mathbf{a} \) by the scalar projection, hence offering both magnitude and direction.
- It generates a new vector aligned with \( \mathbf{a} \).
- Its magnitude is the scalar projection.