Problem 30 - Contemporary Precalculus

Question

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

Step-by-Step Solution

Answer

Question: Find the projections of vector u onto vector v (proju v) and the projection of vector v onto vector u (proju u) if u = 2i - 3j, and v = i + 2j. Answer: The projection of vector u onto vector v (proju v) is -4/5i - 8/5j, and the projection of vector v onto vector u (proju u) is -8/13i + 12/13j.

Step 1: Find the dot product of vectors u and v

To find the dot product of \(\mathbf{u}\) and \(\mathbf{v}\), we can use the formula $$\mathbf{u} \cdot \mathbf{v} = (u_1v_1) + (u_2v_2)$$ In our case, \(u_1=2, u_2=-3, v_1=1,\) and \(v_2= 2\). Then, the dot product is $$\mathbf{u} \cdot \mathbf{v} = (2)(1) + (-3)(2) = 2 - 6 = - 4$$

Step 2: Compute the magnitudes of vectors u and v

To compute the magnitudes of vectors \(\mathbf{u}\) and \(\mathbf{v}\), we can use the formula $$\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2}$$ Applying this to our vectors \(\mathbf{u}\) and \(\mathbf{v}\), we have $$\|\mathbf{u}\| = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$$ $$\|\mathbf{v}\| = \sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$$

Step 3: Compute the projections

Using the formula for projections mentioned in the analysis, we can compute the projections as follows: $$\text{proj}_{\mathbf{v}}(\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} = \frac{-4}{5} (\mathbf{i} + 2\mathbf{j}) = -\frac{4}{5}\mathbf{i}-\frac{8}{5}\mathbf{j}$$ $$\text{proj}_{\mathbf{u}}(\mathbf{v}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|^2} \mathbf{u} = \frac{-4}{13}(2\mathbf{i} - 3\mathbf{j}) = -\frac{8}{13}\mathbf{i}+\frac{12}{13}\mathbf{j}$$ So, the projection of vector \(\mathbf{u}\) onto vector \(\mathbf{v}\) (proju \(v\)) is \(-\frac{4}{5}\mathbf{i}-\frac{8}{5}\mathbf{j}\) and the projection of vector \(\mathbf{v}\) onto vector \(\mathbf{u}\) (proju \(u\)) is \(-\frac{8}{13}\mathbf{i}+\frac{12}{13}\mathbf{j}\).

Key Concepts

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. In simpler terms, it is a way to combine two vectors to get a scalar (a plain number) that represents how "aligned" the two vectors are with each other.
For two vectors \( \mathbf{u} = [u_1, u_2] \) and \( \mathbf{v} = [v_1, v_2] \), the dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated by multiplying the corresponding components of the vectors together and then adding those products:

• First, you multiply the respective components: \( u_1 \times v_1 \) and \( u_2 \times v_2 \).
• Then, you sum the results: \( u_1v_1 + u_2v_2 \).

In the original exercise, we found the dot product of vectors \( \mathbf{u} = 2\mathbf{i} - 3\mathbf{j} \) and \( \mathbf{v} = \mathbf{i} + 2\mathbf{j} \). By substituting the values, the operation becomes \( 2 \times 1 + (-3) \times 2 = 2 - 6 = -4 \), indicating that the vectors are oriented in a way that they have a negative correlation.

Magnitude of a Vector

The magnitude of a vector is essentially its "length". It tells you how long the vector is, regardless of its direction. This concept is crucial because it helps in understanding how big or small the vector is in the coordinate plane.
The formula to calculate the magnitude (length) of a vector \( \mathbf{a} = [a_1, a_2] \) is:

• Using the Pythagorean theorem, it's written as \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \).

For example, for the vector \( \mathbf{u} = 2\mathbf{i} - 3\mathbf{j} \), the magnitude \( \|\mathbf{u}\| \) is calculated as \( \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \).
Similarly, the vector \( \mathbf{v} = \mathbf{i} + 2\mathbf{j} \) has a magnitude \( \|\mathbf{v}\| = \sqrt{1^2 + 2^2} = \sqrt{5} \). Knowing these values is significant in further calculations, such as projections and comparing vector sizes.

Projection Formula

The projection of one vector onto another is a way of "casting" that vector into a new space defined by the second one. This is crucial in many fields like physics and engineering when understanding how forces act in specific directions.
To project vector \( \mathbf{u} \) onto another vector \( \mathbf{v} \), the projection formula is used:

• The formula is \( \text{proj}_{\mathbf{v}}(\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} \).
• Here, \( \mathbf{u} \cdot \mathbf{v} \) is the dot product, and \( \|\mathbf{v}\|^2 \) is the magnitude of \( \mathbf{v} \) squared.

In the exercise, the projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is calculated as:\( \text{proj}_{\mathbf{v}}(\mathbf{u}) = \frac{-4}{5} (\mathbf{i} + 2\mathbf{j}) = -\frac{4}{5}\mathbf{i} - \frac{8}{5}\mathbf{j} \).
This process involves multiplying the scalar \( \frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{v}\|^2} \) by the vector \( \mathbf{v} \) to "reshape" vector \( \mathbf{u} \) into the direction of \( \mathbf{v} \). Observing the projection helps understand how much of vector \( \mathbf{u} \) lies in the direction of vector \( \mathbf{v} \). Similarly, projections can be calculated in the opposite direction, offering insights into both vectors' interactions.