Problem 29 - Essential Calculus Early Transcendentals

Question

\(29-32\) Find the scalar and vector projections of b onto a. $$\mathbf{a}=\langle- 5,12\rangle, \quad \mathbf{b}=\langle 4,6\rangle$$

Step-by-Step Solution

Answer

Scalar projection is 4 and vector projection is approximately \( \langle -1.54, 3.69 \rangle \).

Step 1: Find the Dot Product

The dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2 \). Substituting the values from the vectors, we have:\[ \mathbf{a} \cdot \mathbf{b} = (-5) \cdot 4 + 12 \cdot 6 = -20 + 72 = 52. \]

Step 2: Compute the Magnitude of \( \mathbf{a} \)

The magnitude of vector \( \mathbf{a} \) is calculated using the formula \( ||\mathbf{a}|| = \sqrt{(a_1)^2 + (a_2)^2} \). Thus, we have:\[ ||\mathbf{a}|| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13. \]

Step 3: Calculate the Scalar Projection

The scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by the formula \( \text{scal}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||} \). Substituting the values we found:\[ \text{scal}_{\mathbf{a}} \mathbf{b} = \frac{52}{13} = 4. \]

Step 4: Vector Component Along \( \mathbf{a} \)

To find the vector projection \( \text{proj}_{\mathbf{a}} \mathbf{b} \), we use the formula \( \text{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||^2} \mathbf{a} \). Using the dot product and magnitude squared from earlier steps:Note, \( ||\mathbf{a}||^2 = 169 \), so\[ \text{proj}_{\mathbf{a}} \mathbf{b} = \frac{52}{169} \langle -5, 12 \rangle. \]

Step 5: Solve for Vector Projection

Now perform the multiplication:\[ \text{proj}_{\mathbf{a}} \mathbf{b} = \langle \frac{52}{169} \cdot (-5), \frac{52}{169} \cdot 12 \rangle = \langle -\frac{260}{169}, \frac{624}{169} \rangle. \]Simplifying the fractions gives:\[ \text{proj}_{\mathbf{a}} \mathbf{b} \approx \langle -1.54, 3.69 \rangle. \]

Key Concepts

Dot Product

The dot product is a fundamental concept in vector mathematics. It allows us to determine the degree of alignment between two vectors. To compute the dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \), we use the formula:
\[ \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2. \]
This mathematical expression means we multiply the corresponding components of the vectors and then sum these products.
In our example, vectors \( \mathbf{a} = \langle -5, 12 \rangle \) and \( \mathbf{b} = \langle 4, 6 \rangle \), their dot product is calculated by:

• \( -5 \times 4 = -20 \)
• \( 12 \times 6 = 72 \)

Adding these results, \( -20 + 72 \), gives us \( 52 \).
The dot product is crucial in determining orthogonality or the angle between vectors. A dot product of zero indicates perpendicularity, while a positive or negative dot product indicates an acute or obtuse angle, respectively.

Magnitude of a Vector

The magnitude of a vector is akin to the vector’s length or size. It is calculated using the Pythagorean theorem. For any vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), the magnitude is found with the formula:
\[ ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2}. \]
This involves squaring each component, summing them, and taking the square root of this sum.
In our example, the magnitude of vector \( \mathbf{a} = \langle -5, 12 \rangle \) is:

• \( (-5)^2 = 25 \)
• \( 12^2 = 144 \)

Adding these values results in \( 169 \), and taking the square root yields \( 13 \).
The magnitude will always be a non-negative number and provides insight into how extensive the vector is.

Scalar Projection

The scalar projection is the component of one vector along the direction of another vector. It's essentially a measure of how much of one vector 'lies in the direction' of another.
It is calculated through the formula:
\[ \text{scal}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||}. \]
Which uses the dot product of the vectors and the magnitude of the vector onto which you are projecting.
For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is:

• \( \frac{52}{13} = 4 \)

This means the component of \( \mathbf{b} \) that points in the direction of \( \mathbf{a} \) measures 4 length units in that direction.
_Syntax reminder_: Scalar projections can be positive, indicating a similar direction, or negative, suggesting the vectors point oppositely.

Vector Operations

Vector operations include addition, subtraction, and projections, which combine vectors in different ways to produce new vectors or scalars. For projections, we specifically discuss vector projections:
To find the vector projection, or vector component of \( \mathbf{b} \) along the vector \( \mathbf{a} \), the formula is:
\[ \text{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||^2} \mathbf{a}. \]
This expression projects \( \mathbf{b} \) onto \( \mathbf{a} \) by scaling vector \( \mathbf{a} \) by the scalar \( \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||^2} \).
In our case, the vector projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is:

• First, we calculate \( \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||^2} \), which is \( \frac{52}{169} \).
• Then multiply this scalar by vector \( \mathbf{a} \), giving \[ \langle -\frac{260}{169}, \frac{624}{169} \rangle \approx \langle -1.54, 3.69 \rangle. \]

This tells us how much of vector \( \mathbf{b} \) aligns with vector \( \mathbf{a} \), creating a new vector in \( \mathbf{a} \)'s direction.