Problem 29 - Contemporary Precalculus

Question

Find proju \(v\) and proju u. $$\mathbf{u}=3 \mathbf{i}-5 \mathbf{j}, \mathbf{v}=6 \mathbf{i}+2 \mathbf{j}$$

Step-by-Step Solution

Answer

Question: Calculate the projections $$proj_{\mathbf{u}} \mathbf{v}$$ and $$proj_{\mathbf{v}} \mathbf{u}$$ given the vectors \(\mathbf{u}\) = \(3\mathbf{i} - 5\mathbf{j}\) and \(\mathbf{v}\) = \(6\mathbf{i} + 2\mathbf{j}\). Answer: The projections are $$proj_{\mathbf{u}} \mathbf{v} = \frac{6}{17}\mathbf{i} - \frac{10}{17}\mathbf{j}$$ and $$proj_{\mathbf{v}} \mathbf{u} = \frac{6}{5}\mathbf{i} + \frac{2}{5}\mathbf{j}$$.

Step 1: Write down the vectors and calculate the dot product

First, identify the given vectors \(\mathbf{u}\) = \(3\mathbf{i} - 5\mathbf{j}\) and \(\mathbf{v}\) = \(6\mathbf{i} + 2\mathbf{j}\). Then, calculate their dot product, \(\mathbf{u} \cdot \mathbf{v}\): $$\mathbf{u} \cdot \mathbf{v} = (3\mathbf{i} - 5\mathbf{j}) \cdot (6\mathbf{i} + 2\mathbf{j}) = 3 \cdot 6 + (-5) \cdot 2 = 18 - 10 = 8$$

Step 2: Calculate the magnitude squared of each vector

To compute the magnitude squared of each vector, square its components and add them together: $$\|\mathbf{u}\|^2 = 3^2 + (-5)^2 = 9 + 25 = 34$$ $$\|\mathbf{v}\|^2 = 6^2 + 2^2 = 36 + 4 = 40$$

Step 3: Apply the projection formula for $$proj_{\mathbf{u}} \mathbf{v}$$

According to the projection formula mentioned in the analysis, we have: $$proj_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|^2} \mathbf{u} = \frac{8}{34} (3\mathbf{i} - 5\mathbf{j}) = \frac{2}{17}(3\mathbf{i} - 5\mathbf{j}) = \frac{6}{17}\mathbf{i} - \frac{10}{17}\mathbf{j}$$

Step 4: Apply the projection formula for $$proj_{\mathbf{v}} \mathbf{u}$$

Similarly, using the projection formula for this calculation, we get: $$proj_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} = \frac{8}{40} (6\mathbf{i} + 2\mathbf{j}) = \frac{1}{5}(6\mathbf{i} + 2\mathbf{j}) = \frac{6}{5}\mathbf{i} + \frac{2}{5}\mathbf{j}$$ The projections of the vectors are as follows: $$proj_{\mathbf{u}} \mathbf{v} = \frac{6}{17}\mathbf{i} - \frac{10}{17}\mathbf{j}$$ $$proj_{\mathbf{v}} \mathbf{u} = \frac{6}{5}\mathbf{i} + \frac{2}{5}\mathbf{j}$$

Key Concepts

Dot Product

Understanding the dot product is essential when learning about vector projection. The dot product, also referred to as the scalar product, is a way to multiply two vectors to get a scalar (a single number). It is calculated by multiplying the corresponding components of each vector and then adding up those products.

For vectors \(\mathbf{u}\) and \(\mathbf{v}\), represented as \(\mathbf{u}=a\mathbf{i}+b\mathbf{j}\) and \(\mathbf{v}=c\mathbf{i}+d\mathbf{j}\), the dot product is computed as:\[\mathbf{u} \cdot \mathbf{v} = (a\mathbf{i} + b\mathbf{j}) \cdot (c\mathbf{i} + d\mathbf{j}) = a \cdot c + b \cdot d\]
In the context of our problem, the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) was calculated to be 8.

Vector Magnitude

The magnitude of a vector, often thought of as its 'length', quantifies the distance from the vector's tail to its head when drawn graphically. Mathematically, for a vector \(\mathbf{u}\) with components \(a\mathbf{i}\) and \(b\mathbf{j}\), the magnitude is found using the Pythagorean theorem, which leads to the formula for the magnitude (also known as the Euclidean norm): \(\|\mathbf{u}\| = \sqrt{a^2 + b^2}\).

To find the squared magnitude, which is often used in vector projections, you would simply square each component and add them up:\[\|\mathbf{u}\|^2 = a^2 + b^2\]
In our exercise, the squared magnitudes for \(\mathbf{u}\) and \(\mathbf{v}\) were found to be 34 and 40, respectively.

Projection Formula

The projection of one vector onto another is a way to find the component of one vector that lies along the direction of another vector. The projection formula involves both the dot product and the magnitude of vectors. To project vector \(\mathbf{u}\) onto vector \(\mathbf{v}\), the formula is:\[proj_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v}\]

For our exercise, the projection of \(\mathbf{v}\) onto \(\mathbf{u}\) is calculated using their dot product and the magnitude of \(\mathbf{u}\), resulting in \(\frac{6}{17}\mathbf{i} - \frac{10}{17}\mathbf{j}\). Similarly, projecting \(\mathbf{u}\) onto \(\mathbf{v}\) uses their dot product and the magnitude of \(\mathbf{v}\), giving \(\frac{6}{5}\mathbf{i} + \frac{2}{5}\mathbf{j}\).
This operation has practical applications in fields like physics, where it helps in decomposing forces into components along specified directions.