Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each of the given projections if \(\mathbf{u}=\mathbf{i}+2 \mathbf{j}, \mathbf{v}=2 \mathbf{i}-\mathbf{j}\), and \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}\). \(\operatorname{proj}_{\mathbf{u}} \mathbf{v}\)

Step-by-Step Solution

Answer

The projection of \( \mathbf{v} \) onto \( \mathbf{u} \) is \( \mathbf{0}. \)

Step 1: Find the Dot Product of u and v

The dot product of two vectors \( \mathbf{a} = a_1\mathbf{i}+a_2\mathbf{j}\) and \( \mathbf{b} = b_1\mathbf{i}+b_2\mathbf{j}\) is calculated as \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).For \( \mathbf{u} = \mathbf{i} + 2\mathbf{j} \) and \( \mathbf{v} = 2\mathbf{i} - \mathbf{j} \), the dot product is:\(\mathbf{u} \cdot \mathbf{v} = (1)(2) + (2)(-1) = 2 - 2 = 0. \)Thus, \( \mathbf{u} \cdot \mathbf{v} = 0. \)

Step 2: Find the Magnitude Squared of u

The formula for the magnitude squared of a vector \( \mathbf{a} = a_1\mathbf{i}+a_2\mathbf{j} \) is \( ||\mathbf{a}||^2 = a_1^2 + a_2^2 \).Given \( \mathbf{u} = \mathbf{i} + 2\mathbf{j} \), compute the magnitude squared:\(||\mathbf{u}||^2 = (1)^2 + (2)^2 = 1 + 4 = 5.\)

Step 3: Calculate the Projection of v onto u

The projection of \( \mathbf{b} \) onto \( \mathbf{a} \) is given by the formula:\(\operatorname{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||^2} \mathbf{a}.\)Using \( \mathbf{u} \cdot \mathbf{v} = 0 \) and \( ||\mathbf{u}||^2 = 5 \), the projection is:\(\operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{0}{5} \mathbf{u} = 0\mathbf{u} = \mathbf{0}.\)Hence, the projection is the zero vector, \( \mathbf{0}. \)

Key Concepts

Dot Product

The dot product is a fundamental concept in vector algebra. It focuses on two vectors and gives us a single scalar value as the result, showcasing how much one vector goes in the direction of another. For two-dimensional vectors, such as \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} \), we calculate the dot product using:

• Multiply the corresponding components of each vector.
• Sum these products to get the final scalar value.

This is represented in the formula:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]In practical terms, if vectors are aligned, the dot product is maximized, and when they are orthogonal, the dot product is zero. In the given exercise with \( \mathbf{u} = \mathbf{i} + 2\mathbf{j} \) and \( \mathbf{v} = 2\mathbf{i} - \mathbf{j} \), the dot product equates to zero, indicating the vectors are perpendicular. This orthogonality signifies no projection of one vector onto another.

Vector Magnitude

Understanding a vector's magnitude is like measuring the length of a vector. This is crucial when calculating projections, as the vector's length impacts the result. To find the magnitude of a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \), compute the square root of the sum of the squares of its components:

• Square each component of the vector.
• Add these squares together to get the magnitude squared.
• Calculate the square root of this sum for the magnitude.

Mathematically, this looks like:\[ ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2} \]The problem uses the squared magnitude \( ||\mathbf{u}||^2 = 5 \) to simplify calculations for projections. The magnitude helps us assess the scale at which one vector projects onto another.

Projection Formula

The projection formula allows us to find a vector's shadow onto another vector. This calculation tells us how one vector is "projected" or "mapped" along the direction of another. The formula for projecting \( \mathbf{b} \) onto \( \mathbf{a} \) is:\[ \operatorname{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||^2} \mathbf{a} \]With this:

• We first find the dot product between \( \mathbf{a} \) and \( \mathbf{b} \).
• Then, divide this dot product by the magnitude squared of \( \mathbf{a} \).

The result is then scaled in the direction of \( \mathbf{a} \). In our exercise, the projection results in the zero vector \( \mathbf{0} \) because the dot product \( \mathbf{u} \cdot \mathbf{v} \) was zero. This zero result demonstrates that when vectors are perpendicular, no part of one vector lies along the other.