Problem 20 - Advanced Engineering Mathematics

Question

Determine a scalar \(c\) so that the angle between \(\mathbf{a}=\mathbf{i}+c \mathbf{j}\) and \(\mathbf{b}=\mathbf{i}+\mathbf{j}\) is \(45^{\circ}\).

Step-by-Step Solution

Answer

The scalar \( c \) is 0.

Step 1: Understanding the Dot Product Formula

To find the angle between two vectors, we use the dot product formula: \[ \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos(\theta) \] where \( \mathbf{a}\) and \( \mathbf{b} \) are vectors, and \( \theta \) is the angle between them. In this case, \( \theta = 45^{\circ}\), so \(\cos(\theta) = \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \).

Step 2: Compute Dot Product of Vectors

Calculate the dot product of \( \mathbf{a} = \mathbf{i} + c \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} + \mathbf{j} \): \[ \mathbf{a} \cdot \mathbf{b} = (1)(1) + (c)(1) = 1 + c \]

Step 3: Find the Magnitudes of the Vectors

Compute the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \):\[ \| \mathbf{a} \| = \sqrt{1^2 + c^2} = \sqrt{1 + c^2} \] \[ \| \mathbf{b} \| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

Step 4: Apply the Dot Product and Magnitudes to the Cosine Formula

Plug the dot product and magnitudes into the cosine formula: \[ 1 + c = \frac{\sqrt{2}}{2} \times \sqrt{1 + c^2} \times \sqrt{2} \] which simplifies to: \[ 1+c = \sqrt{1+c^2} \]

Step 5: Solve for c

Square both sides to eliminate the square root: \[ (1+c)^2 = 1+c^2 \] which simplifies to: \[ 1 + 2c + c^2 = 1 + c^2 \] Cancel \(1 + c^2\) from both sides: \[ 2c = 0 \] Thus, \( c = 0 \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It is often used to find the angle between two vectors. To compute the dot product of two vectors, you multiply their corresponding components and add the results together.
For example, given vectors \( \mathbf{a} = \mathbf{i} + c \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} + \mathbf{j} \), the dot product is calculated as follows:

• Multiply the \( \mathbf{i} \) components: \( 1 \times 1 \).
• Multiply the \( \mathbf{j} \) components: \( c \times 1 \).
• Add these products: \( 1 + c \).

The result gives us the dot product \( \mathbf{a} \cdot \mathbf{b} = 1 + c \). This value is crucial for determining angles between vectors when combined with vector magnitudes.

Magnitude of Vectors

The magnitude of a vector is its length in geometric terms. You can think of it as the straight-line distance from the origin to the point defined by the vector.
To calculate the magnitude of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), use the formula:

• \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \).

For vector \( \mathbf{a} = \mathbf{i} + c \mathbf{j} \), the magnitude is \( \| \mathbf{a} \| = \sqrt{1 + c^2} \).
For vector \( \mathbf{b} = \mathbf{i} + \mathbf{j} \), it is \( \| \mathbf{b} \| = \sqrt{2} \).
The magnitude helps in normalizing a vector or finding the angle between vectors when used with the dot product.

Angle Between Vectors

The angle between two vectors can tell us about their directional relationship. Calculating this angle involves both the dot product and the magnitudes of the vectors.
Once you have both, you can use the formula for the angle \( \theta \):

• \( \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

This formula essentially expresses the cosine of the angle as a ratio of the two previously computed quantities.
In our example, knowing the angle is \( 45^{\circ} \) helps set \( \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \), allowing us to solve for other unknowns, such as the scalar \( c \) in the problem. This extends our understanding from simple calculation to practical application.

Cosine Formula

The cosine formula in vector mathematics helps calculate the angle between vectors when combined with dot products and magnitudes. The formula, as described, is:

• \( \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos(\theta) \)

This allows us to solve problems involving angles by rearranging the formula when necessary.
Given the problem, we substitute known values, including the specified angle of \( 45^{\circ} \) which gives \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \).
To find the scalar \( c \), we align it through simplifying expressions such as \( \sqrt{1+c^2} \) and equating \( 1+c = \sqrt{1+c^2} \).
Through algebra, it allows us to solve \( c = 0 \), showing the utility and power of the cosine formula in vector calculations.