Problem 29 - Precalculus Mathematics for Calculus

Question

Calculate proj, u. (b) Resolve u into \(u_{1}\) and \(u_{2}\), where \(\mathbf{u}_{1}\) is parallel to \(\mathbf{v}\) and \(\mathbf{u}_{2}\) is orthogonal to \(\mathbf{v}\). $$\mathbf{u}=\langle- 2,4\rangle, \quad \mathbf{v}=\langle 1,1\rangle$$

Step-by-Step Solution

Answer

\(\mathbf{u}_1 = \langle 1, 1 \rangle\) and \(\mathbf{u}_2 = \langle -3, 3 \rangle\).

Step 1: Calculate the Dot Product of u and v

To begin, calculate the dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}\). The dot product is given by:\[\mathbf{u} \cdot \mathbf{v} = (-2)(1) + (4)(1) = -2 + 4 = 2\]

Step 2: Calculate the Magnitude of v

Next, find the magnitude of vector \(\mathbf{v}\). This can be calculated as follows:\[||\mathbf{v}|| = \sqrt{(1)^2 + (1)^2} = \sqrt{2}\]

Step 3: Calculate the Projection of u onto v

The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is given by the formula:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{v}||^2} \mathbf{v} = \frac{2}{2} \mathbf{v} = \mathbf{v} = \langle 1, 1 \rangle\]

Step 4: Resolve u into u1 and u2

Now, resolve \(\mathbf{u}\) into \(\mathbf{u}_1\) and \(\mathbf{u}_2\). We already found \(\mathbf{u}_1 = \text{proj}_{\mathbf{v}} \mathbf{u} = \langle 1, 1 \rangle\). To find \(\mathbf{u}_2\), compute:\[\mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 = \langle -2, 4 \rangle - \langle 1, 1 \rangle = \langle -3, 3 \rangle\]

Step 5: Verify Orthogonality

Finally, verify that \(\mathbf{u}_2\) is orthogonal to \(\mathbf{v}\) by checking their dot product:\[\mathbf{u}_2 \cdot \mathbf{v} = (-3)(1) + (3)(1) = -3 + 3 = 0\]A dot product of zero confirms orthogonality.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. The formula for the dot product of two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\) is: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \] Essentially, it multiplies corresponding components of the vectors and adds them together.

Let's break down its significance:

• Resulting Value: The result is a scalar, not a vector.
• Angle Interpretation: The dot product relates to the cosine of the angle \(\theta\) between \(\mathbf{a}\) and \(\mathbf{b}\). If the dot product is zero, the vectors are orthogonal (perpendicular to each other).
• Applications: Used in projections, determining angles, and physics problems involving work.

In our example, the dot product of \(\mathbf{u} = \langle -2, 4 \rangle\) and \(\mathbf{v} = \langle 1, 1 \rangle\) is calculated as \(-2\times1 + 4\times1 = 2\). This scalar value is crucial for determining how aligned \(\mathbf{u}\) and \(\mathbf{v}\) are.

Magnitude of a Vector

The magnitude of a vector is its length in space; think of it as the size or extent of a vector. For a vector \(\mathbf{a} = \langle a_1, a_2 \rangle\), its magnitude is given by: \[ ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2} \] This measures the vector's distance from the origin to its endpoint in a coordinate system.

Why is vector magnitude important?

• Normalization: It allows us to create unit vectors by dividing the vector by its magnitude, thus scaling the vector to have a length of one.
• Understanding Vector Relations: Helps compare vector sizes and determine proportions in vector projections.
• Physical Interpretation: In physics, it represents quantities like speed, force, or any vector quantity's strength.

In our solution, the magnitude of \(\mathbf{v} = \langle 1, 1 \rangle\) is calculated as \(\sqrt{1^2 + 1^2} = \sqrt{2}\). This value plays a role in determining the projection of \(\mathbf{u}\) onto \(\mathbf{v}\).

Orthogonal Vectors

Orthogonal vectors are vectors that meet at right angles, meaning they are perpendicular to each other. The quick way to check if two vectors are orthogonal is to find their dot product. If the dot product is zero, the vectors are orthogonal.

Why orthogonality matters:

• Geometric Interpretation: It signifies a lack of projection of one vector on the other, indicating they influence each other minimally if considered as forces or directions.
• Applications: Used in graphics and engineering, for determining independence of vectors, and in orthogonal decompositions.
• Solving Problems: Essential in breaking down vectors into components such as in linear algebra and trigonometry.

In our example, when resolving \(\mathbf{u}\) into \(\mathbf{u}_1\) and \(\mathbf{u}_2\), we found that \(\mathbf{u}_2 = \langle -3, 3 \rangle\) is orthogonal to \(\mathbf{v}\). The dot product of \(\mathbf{u}_2\) and \(\mathbf{v}\) was calculated as \(-3\times1 + 3\times1 = 0\), confirming their orthogonality.