Problem 26 - Algebra and Trigonometry

Question

23-28 (a) Calculate proj, \(\mathbf{u} .\) (b) Resolve \(\mathbf{u}\) into \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) where \(\mathbf{u}_{1}\) is parallel to \(\mathbf{v}\) and \(\mathbf{u}_{2}\) is orthogonal to \(\mathbf{v}\). $$\mathbf{u}=\langle 11,3\rangle, \quad \mathbf{v}=\langle- 3,-2\rangle$$

Step-by-Step Solution

Answer

\( \text{proj}_{\mathbf{v}} \mathbf{u} = \langle 9, 6 \rangle \), \( \mathbf{u}_2 = \langle 2, -3 \rangle \)

Step 1: Calculate Dot Product

To find the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), we first calculate the dot product of \( \mathbf{u} \) and \( \mathbf{v} \). The dot product is given by \( \mathbf{u} \cdot \mathbf{v} = 11(-3) + 3(-2) = -33 - 6 = -39 \).

Step 2: Find Magnitude of \( \mathbf{v} \)

Next, we compute the magnitude of \( \mathbf{v} \). The magnitude is given by \( \| \mathbf{v} \| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \).

Step 3: Calculate Projection of \( \mathbf{u} \) onto \( \mathbf{v} \)

The projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is calculated using the formula:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} \]Substituting the known values, we have\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{-39}{\sqrt{13}^2} \langle -3, -2 \rangle = \frac{-39}{13} \langle -3, -2 \rangle = -3 \langle -3, -2 \rangle = \langle 9, 6 \rangle \]

Step 4: Find \( \mathbf{u}_2 \) (Component Orthogonal to \( \mathbf{v} \))

Now, find \( \mathbf{u}_2 \), the component of \( \mathbf{u} \) that is orthogonal to \( \mathbf{v} \). This is given by:\( \mathbf{u}_2 = \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u} \).Substituting the values gives:\( \mathbf{u}_2 = \langle 11, 3 \rangle - \langle 9, 6 \rangle = \langle 2, -3 \rangle \).

Step 5: Verify Orthogonality

To verify that \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \), compute their dot product:\( \mathbf{u}_2 \cdot \mathbf{v} = 2(-3) + (-3)(-2) = -6 + 6 = 0 \).Since the dot product is zero, \( \mathbf{u}_2 \) is indeed orthogonal to \( \mathbf{v} \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It's a way to multiply two vectors resulting in a scalar (a single number) rather than another vector. In simple terms, the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2\rangle \) and \( \mathbf{b} = \langle b_1, b_2\rangle \) is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 \)

For our vectors \( \mathbf{u} = \langle 11, 3 \rangle \) and \( \mathbf{v} = \langle -3, -2 \rangle \), their dot product is \(-39\).
This calculation shows that there's some angle between them. Remember, a dot product of zero indicates that vectors are orthogonal (i.e., at a right angle), but a non-zero dot product means the vectors are not perpendicular to each other.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at a right angle. In mathematical terms, they are perpendicular. The key property of orthogonal vectors is that their dot product is always zero. This is because the cosine of the angle \(90^{\circ}\) between them is zero.
In vector resolution, determining orthogonal components is essential. For example, if you have a vector \( \mathbf{u} \), you can resolve it into two parts:

• \( \mathbf{u}_1 \), which is parallel to another vector \( \mathbf{v} \)
• \( \mathbf{u}_2 \), which is orthogonal to \( \mathbf{v} \)

In our exercise, after finding the projection, the orthogonal vector \( \mathbf{u}_2 = \langle 2, -3 \rangle \) is calculated by subtracting the projection from \( \mathbf{u} \).
This ensures \( \mathbf{u}_2 \cdot \mathbf{v} = 0 \), verifying orthogonality, as shown by the dot product calculation yielding zero.

Vector Magnitude

The magnitude of a vector provides a measure of its length or size. Using components, the magnitude \( \| \mathbf{v} \| \) of vector \( \mathbf{v} \) can be calculated through the Pythagorean theorem:

• \( \| \mathbf{v} \| = \sqrt{b_1^2 + b_2^2} \)

This gives a scalar quantity that represents how large or long the vector is.
Applying this to \( \mathbf{v} = \langle -3, -2 \rangle \), we find \( \| \mathbf{v} \| = \sqrt{13} \). This step is crucial when calculating projections, as seen in the formula for the projection of \( \mathbf{u} \) onto \( \mathbf{v} \):
\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} \]
This formula relies on knowing the magnitude to accurately scale \( \mathbf{v} \) in the direction of \( \mathbf{u} \). Understanding vector magnitude helps us visualize the vector's direction and length in space.