Problem 8 - Algebra and Trigonometry

Question

Find (a) \(\mathbf{u} \cdot \mathbf{v}\) and (b) the angle between \(\mathbf{u}\) and \(\mathbf{v}\) to the nearest degree. $$ \mathbf{u}=\langle- 6,6\rangle, \quad \mathbf{v}=\langle 1,-1\rangle $$

Step-by-Step Solution

Answer

(a) \(-12\); (b) \(180^\circ\).

Step 1: Calculate the Dot Product

The dot product of two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \) is given by the formula \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). For the vectors \( \mathbf{u} = \langle -6, 6 \rangle \) and \( \mathbf{v} = \langle 1, -1 \rangle \), calculate:\[(-6)(1) + (6)(-1) = -6 - 6 = -12.\]The dot product \( \mathbf{u} \cdot \mathbf{v} \) is \(-12\).

Step 2: Calculate the Magnitudes of the Vectors

To find the angle between the vectors, we first need their magnitudes. The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is given by \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \).For \( \mathbf{u} = \langle -6, 6 \rangle \):\[\| \mathbf{u} \| = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}.\]For \( \mathbf{v} = \langle 1, -1 \rangle \):\[\| \mathbf{v} \| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}.\]

Step 3: Calculate the Cosine of the Angle Between Vectors

The cosine of the angle \( \theta \) between two vectors can be found using the dot product and magnitudes: \[\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \| \mathbf{v} \|} = \frac{-12}{6\sqrt{2} \times \sqrt{2}} = \frac{-12}{12} = -1.\]

Step 4: Find the Angle Using Inverse Cosine

To find the angle \( \theta \), take the inverse cosine (\( \cos^{-1} \)) of \(-1\). The angle whose cosine is \(-1\) is \(180^\circ\).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector calculus. It allows you to multiply two vectors, resulting in a scalar. To understand this better, let's consider two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \). The dot product, denoted by \( \mathbf{u} \cdot \mathbf{v} \), is calculated using the formula:

• Multiply each component of the vectors: \( u_1v_1 + u_2v_2 \).
• In our case: \((-6)(1) + (6)(-1) = -6 - 6 = -12\).

This gives us a single number, \(-12\), which tells us about the projection of one vector onto another. A positive dot product implies the vectors are somewhat aligned, while a negative dot product means they are pointing in opposite directions. When the dot product is zero, it indicates the vectors are perpendicular.

Vector Magnitude

The magnitude of a vector is a measure of its length or size. It's like finding the length of a line segment represented by the vector. To find the magnitude of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), we use the formula:

• \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \).

Let's calculate:

• For \( \mathbf{u} = \langle -6, 6 \rangle \), the magnitude is \( \sqrt{(-6)^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \).
• For \( \mathbf{v} = \langle 1, -1 \rangle \), the magnitude is \( \sqrt{1^2 + (-1)^2} = \sqrt{2} \).

Finding the magnitude turns vectors into simpler scalar values that represent their length.

Angle Between Vectors

The angle between two vectors is key in understanding how they relate geometrically. To find the angle, we first use the dot product and magnitudes of the vectors. The formula that connects them is the cosine rule:

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

In our example:

• \( \cos \theta = \frac{-12}{6\sqrt{2} \cdot \sqrt{2}} = \frac{-12}{12} = -1 \).
• This means \( \theta \) is \( 180^\circ \) because the inverse cosine of \(-1\) is \( 180^\circ \).

Thus, \( \mathbf{u} \) and \( \mathbf{v} \) are pointing in exactly opposite directions. Knowing the angle provides insight into the spatial relationship and alignment of the vectors.