Problem 6 - Precalculus Mathematics for Calculus

Question

Find \((a) u \cdot v\) and \((b)\) the angle between \(u\) and \(v\) to the nearest degree. $$\mathbf{u}=\mathbf{i}+\sqrt{3} \mathbf{j}, \quad \mathbf{v}=-\sqrt{3} \mathbf{i}+\mathbf{j}$$

Step-by-Step Solution

Answer

(a) \( u \cdot v = 0 \); (b) angle is 90°.

Step 1: Write the Vectors in Component Form

Vectors \( \mathbf{u} \) and \( \mathbf{v} \) can be written in component form as \( \mathbf{u} = \langle 1, \sqrt{3} \rangle \) and \( \mathbf{v} = \langle -\sqrt{3}, 1 \rangle \).

Step 2: Calculate Dot Product \( u \cdot v \)

To find the dot product, use the formula for dot product of two vectors: \( u \cdot v = \mathbf{u}_1 \cdot \mathbf{v}_1 + \mathbf{u}_2 \cdot \mathbf{v}_2 \). Substituting the components, we get: \( 1 \cdot (-\sqrt{3}) + \sqrt{3} \cdot 1 = -\sqrt{3} + \sqrt{3} = 0 \). Thus, \( u \cdot v = 0 \).

Step 3: Use Dot Product to Find Cosine of Angle Between Vectors

Use the dot product to find the cosine of the angle between two vectors: \( u \cdot v = \|u\| \|v\| \cos \theta \). Since \( u \cdot v = 0 \), the angle \( \theta \) satisfies \( \cos \theta = 0 \). This means \( \theta = 90^\circ \).

Step 4: Find the Magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \)

The magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \) is given by \( \|v\| = \sqrt{a^2 + b^2} \). For \( \mathbf{u} = \langle 1, \sqrt{3} \rangle \), \( \|u\| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2 \). For \( \mathbf{v} = \langle -\sqrt{3}, 1 \rangle \), \( \|v\| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \).

Step 5: Conclusion

With \( \cos \theta = 0 \), the angle \( \theta \) between \( \mathbf{u} \) and \( \mathbf{v} \) is \( 90^\circ \). The provided solution and magnitudes verify perpendicular vectors.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector calculus. It provides a way to multiply two vectors together to get a scalar. To calculate the dot product, you multiply the corresponding components of the vectors and then sum those products.
For example, given two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), the dot product is computed as:

• \( u \cdot v = u_1 \cdot v_1 + u_2 \cdot v_2 \)

In the problem, we computed the dot product of \( \mathbf{u} = \langle 1, \sqrt{3} \rangle \) and \( \mathbf{v} = \langle -\sqrt{3}, 1 \rangle \).

This resulted in a computation of:

• \( 1 \cdot (-\sqrt{3}) + \sqrt{3} \cdot 1 = -\sqrt{3} + \sqrt{3} = 0 \)

Since the dot product here is zero, this indicates that the vectors are perpendicular to each other.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length, is crucial for understanding the size of a vector. To find the magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \), you use the formula:

• \( \|v\| = \sqrt{a^2 + b^2} \)

This formula reflects the Pythagorean theorem, where the magnitude is the hypotenuse of a right triangle formed by the vector's components.
For vector \( \mathbf{u} = \langle 1, \sqrt{3} \rangle \), we computed:

• \( \|u\| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2 \)

Similarly, for vector \( \mathbf{v} = \langle -\sqrt{3}, 1 \rangle \):

• \( \|v\| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \)

Both vectors have magnitudes of 2, which confirms they are of equal length. Understanding vector magnitude helps in finding not only the size but also in applications such as normalization.

Angle Between Vectors

Calculating the angle between two vectors involves understanding how to use both the dot product and magnitudes. The cosine of the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is determined by:

• \( \cos \theta = \frac{u \cdot v}{\|u\| \|v\|} \)

In simpler terms, this arranges the relationship of their orientations.
It is key to note that when the dot product \( u \cdot v = 0 \), the cosine value is also zero, which implies an angle of \( 90^\circ \).
This calculation highlight was evident in our problem, where \( \mathbf{u} \) and \( \mathbf{v} \) resulted in \( \cos \theta = 0 \).
Thus, the vectors are perpendicular. If a different result occurs, you adjust the conclusion about their interaction based on the cosine value.