Problem 32 - Essential Calculus Early Transcendentals

Question

\(29-32\) Find the scalar and vector projections of b onto a. $$\mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \quad \mathbf{b}=\mathbf{i}-\mathbf{j}+\mathbf{k}$$

Step-by-Step Solution

Answer

Scalar projection: \( \frac{1}{\sqrt{3}} \); Vector projection: \( \frac{1}{3}(\mathbf{i} + \mathbf{j} + \mathbf{k}) \).

Step 1: Compute the Dot Product

The vector dot product is calculated using the formula: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). Here, \( \mathbf{a} = \langle 1, 1, 1 \rangle \) and \( \mathbf{b} = \langle 1, -1, 1 \rangle \). Therefore, \( \mathbf{a} \cdot \mathbf{b} = 1 \times 1 + 1 \times (-1) + 1 \times 1 = 1 - 1 + 1 = 1 \).

Step 2: Compute the Magnitude of Vector a

The magnitude of \( \mathbf{a} \) is \( \| \mathbf{a} \| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).

Step 3: Find the Scalar Projection of b onto a

The scalar projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by the formula: \( \text{scal}_a(\mathbf{b}) = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \|} \). Substituting the values gives \( \frac{1}{\sqrt{3}} \).

Step 4: Compute the Vector Projection of b onto a

The vector projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by the formula: \( \text{proj}_a(\mathbf{b}) = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a} \). Here, \( \mathbf{a} \cdot \mathbf{a} = \| \mathbf{a} \|^2 = 3 \). Thus, \( \text{proj}_a(\mathbf{b}) = \frac{1}{3} \mathbf{a} = \frac{1}{3}(\mathbf{i} + \mathbf{j} + \mathbf{k}) \).

Key Concepts

Scalar Projections

Scalar projections help us understand how one vector extends along the direction of another. In simpler terms, it measures the length of the "shadow" cast by one vector onto another vector when light shines from above. To compute the scalar projection of vector \( \mathbf{b} \) onto vector \( \mathbf{a} \), we calculate:

• The Dot Product: First, calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \), which combines the two vectors' magnitudes and directions. For vectors \( \mathbf{a} = \langle 1, 1, 1 \rangle \) and \( \mathbf{b} = \langle 1, -1, 1 \rangle \), this value is 1, as shown by the solution's step.


• Division by \( \mathbf{a} \'s \) magnitude: Then, divide the dot product by the magnitude of vector \( \mathbf{a} \), \( \| \mathbf{a} \| \). It is calculated as \( \sqrt{3} \).
  Thus, the scalar projection \( \text{scal}_a(\mathbf{b}) \) becomes \( \frac{1}{\sqrt{3}} \).

Scalar projections provide insight into how much of vector \( \mathbf{b} \) is in the direction of vector \( \mathbf{a} \). This can determine alignment or force decomposition in physics.

Dot Product

The dot product is a fundamental operation in vector mathematics. It describes how much one vector goes in the direction of another. Here's how it works:

• Understanding Components: Each vector has components. For vector \( \mathbf{a} = \langle 1, 1, 1 \rangle \), its components are along each unit axis \( (i, j, k) \). The same concept applies to vector \( \mathbf{b} = \langle 1, -1, 1 \rangle \).


• Calculation: The formula for the dot product \( \mathbf{a} \cdot \mathbf{b} \) is a sum of products of corresponding components:
  \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = 1\times1 + 1\times(-1) + 1\times1 = 1 \]

The dot product can tell us about the relationship between two vectors:

• When it's positive, vectors point in a generally similar direction.
• When zero, vectors are perpendicular.
• When negative, vectors mostly oppose each other.

Understanding the dot product helps with projections, determining angles between vectors, and being the basis for different vector operations.

Magnitude of a Vector

Understanding the magnitude of a vector is crucial for grasping vectors' sizes in different directions. The magnitude, also called the length, indicates how long the vector is and is expressed as \( \| \mathbf{v} \| \). Here's how you calculate it:

• Formula: For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), its magnitude is calculated using the formula:
  \[ \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]


• Example: For vector \( \mathbf{a} = \langle 1, 1, 1 \rangle \), the magnitude computes to \( \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).

This measure is pivotal when determining the scalar and vector projections. The magnitude plays a critical role in normalizing vectors—converting them into unit vectors—allowing for vector comparisons regardless of their original lengths.