Problem 30 - Essential Calculus Early Transcendentals

Question

\(29-32\) Find the scalar and vector projections of b onto a. $$\mathbf{a}=\langle 1,4\rangle, \quad \mathbf{b}=\langle 2,3\rangle$$

Step-by-Step Solution

Answer

The scalar projection is \(\frac{14}{\sqrt{17}}\), and the vector projection is \(\langle \frac{14}{17}, \frac{56}{17} \rangle\).

Step 1: Calculate the Dot Product

First, determine the dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\). The dot product formula is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \). For \( \mathbf{a} = \langle 1, 4 \rangle \) and \( \mathbf{b} = \langle 2, 3 \rangle \), the calculation is: \(\mathbf{a} \cdot \mathbf{b} = 1 \cdot 2 + 4 \cdot 3 = 2 + 12 = 14\).

Step 2: Calculate the Magnitude of Vector a

Next, compute the magnitude (length) of \(\mathbf{a}\) using the formula for the magnitude of a vector: \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \). For \(\mathbf{a} = \langle 1, 4 \rangle\), this is \(\|\mathbf{a}\| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}\).

Step 3: Compute the Scalar Projection

The scalar projection of \(\mathbf{b}\) onto \(\mathbf{a}\) is given by \( \mathrm{comp}_a(b) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|} \). Substitute the values we have: \( \mathrm{comp}_a(b) = \frac{14}{\sqrt{17}}\).

Step 4: Compute the Vector Projection

To find the vector projection of \(\mathbf{b}\) onto \(\mathbf{a}\), use \( \mathrm{proj}_a(b) = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2} \right) \mathbf{a} \). Calculate \( \|\mathbf{a}\|^2 \) first: \( \|\mathbf{a}\|^2 = 17 \). Thus, \( \mathrm{proj}_a(b) = \left( \frac{14}{17} \right) \langle 1, 4 \rangle = \langle \frac{14}{17}, \frac{56}{17} \rangle \).

Key Concepts

Scalar Projections

The concept of scalar projection is fundamental to understanding how much of one vector stretches out in the direction of another. Imagine you have two vectors:

• \(\mathbf{a}\)
• \(\mathbf{b}\)

The scalar projection essentially tells "how much" of \(\mathbf{b}\) falls along \(\mathbf{a}\). It's a measure of length in the direction of \(\mathbf{a}\) but without any direction itself.
To calculate the scalar projection of \(\mathbf{b}\) onto \(\mathbf{a}\), use the formula:\[ \text{comp}_a(b) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|} \]The dot product (\(\mathbf{a} \cdot \mathbf{b}\)) sits in the numerator and represents a scalar value reflecting the total magnitude when \(\mathbf{b}\) points directly in the direction of \(\mathbf{a}\). In our exercise, for vectors \(\mathbf{a} = \langle 1, 4 \rangle\) and \(\mathbf{b} = \langle 2, 3 \rangle\), the scalar projection is \(\frac{14}{\sqrt{17}}\). This result signifies that if you were to 'flatten' \(\mathbf{b}\) onto \(\mathbf{a}\), it would extend by \(\frac{14}{\sqrt{17}}\) units along \(\mathbf{a}\).

Dot Product

The dot product is an essential operation in vector mathematics, serving as a bridge between two vectors to consolidate their interactions. When you compute the dot product

• \(\mathbf{a} \cdot \mathbf{b}\)

you're essentially calculating how much \(\mathbf{b}\) contributes to the direction of \(\mathbf{a}\). This measure considers both magnitude and alignment.
The formula for the dot product is:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n \]where each respective component of \(\mathbf{a}\) and \(\mathbf{b}\) gets multiplied together. It's like multiplying two lists of numbers—aligning elements and summing the results. In the current exercise, for vectors \(\mathbf{a} = \langle 1, 4 \rangle\) and \(\mathbf{b} = \langle 2, 3 \rangle\), the dot product informs us that the summation
(\(1 \times 2 + 4 \times 3 = 14\))represents both vectors' overlapping influence on each other.
If the dot product is zero, it means these vectors are perpendicular and have no shared directional influence.

Magnitude of a Vector

Understanding the magnitude of a vector is like grasping how long a line segment is in geometry. The magnitude, also known as the length, quantifies how far a vector stretches in space. For a vector \(\mathbf{a} = \langle a_1, a_2 \rangle\), the magnitude is found using:\[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \]This equation is an adaptation of the Pythagorean theorem. It computes the hypotenuse (longest side) of a right triangle formed between the origin and the vector's endpoint.
In our scenario, vector \(\mathbf{a} = \langle 1, 4 \rangle\), its magnitude is:\(\|\mathbf{a}\| = \sqrt{1^2 + 4^2} = \sqrt{17}\).
This means that \(\mathbf{a}\) reaches for approximately 4.123 units in the two-dimensional plane since \(\sqrt{17} \approx 4.123\). The magnitude helps us comprehend not only the size of a vector, but also supports other computations like normalizing or projecting vectors onto one another.