Problem 79 - Precalculus with Limits

Question

Using the Law of Cosines In Exercises 79 and 80 , use the Law of Cosines to find the angle \(\alpha\) between the vectors. (Assume \(0^{\circ} \leq \alpha \leq 180^{\circ} . )\) $$\mathbf{v}=\mathbf{i}+\mathbf{j}, \quad \mathbf{w}=2 \mathbf{i}-2 \mathbf{j}$$

Step-by-Step Solution

Answer

The angle between the vectors \(\mathbf{v}\) and \(\mathbf{w}\) is \(90^{\circ}\).

Step 1: Compute the dot product of the vectors

The dot product of two vectors \(\mathbf{v}=(v_1, v_2)\) and \(\mathbf{w}=(w_1, w_2)\) is given by \(v_1w_1 + v_2w_2\). For the given vectors, this computation gives \((1)(2)+ (1)(-2) = 2-2=0 \)

Step 2: Compute the magnitudes of the vectors

The magnitude of a vector \(\mathbf{v}=(v_1, v_2)\) is given by \(\sqrt[]{v_1^2 + v_2^2}\). For \(\mathbf{v}\), this gives \(\sqrt[]{1^2 + 1^2} = \sqrt[]{2}\), and for \(\mathbf{w}\), this gives \(\sqrt[]{2^2 + (-2)^2} = 2\sqrt[]{2}\)

Step 3: Compute the cosine of the angle

According to the Law of Cosines, the cosine of the angle \(\alpha\) between the vectors is given by \(\cos \alpha = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| ||\mathbf{w}||}\). Substituting the computed values gives \(\cos \alpha = \frac{0}{2} = 0\)

Step 4: Compute the angle

Finally, the angle \(\alpha\) is obtained by computing the arc cosine of the computed cosine value. Therefore, \(\alpha = \arccos(0) = 90^{\circ}\).

Key Concepts

Vector Magnitude

Understanding vector magnitude is essential for delving into vector-related problems in physics and mathematics. A vector represents a quantity possessing both magnitude (size) and direction. To visualize it, imagine an arrow pointing from one point to another. The length of this arrow measures the vector's magnitude.

Mathematically, the magnitude of a two-dimensional vector \( \mathbf{v} = (v_1, v_2) \) is calculated as \( ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2} \). For the given example \( \mathbf{v} = \mathbf{i} + \mathbf{j} \), the magnitude would be \( ||\mathbf{v}|| = \sqrt{1^2 + 1^2} = \sqrt{2} \).

Having a solid grasp of how to compute vector magnitude allows one to analyze vector quantities in various scientific and engineering contexts, such as determining the length of a displacement vector or the strength of a force.

Dot Product

When dealing with vectors, the dot product is a definitive way to express the product of their magnitudes and the cosine of the angle between them. It's a scalar, which means it's a single number as opposed to a vector that has both magnitude and direction.

The formula for the dot product of two vectors \( \mathbf{v} \) and \( \mathbf{w} \) in two dimensions is \( \mathbf{v} \) \( \cdot \) \( \mathbf{w} = v_1w_1 + v_2w_2 \). From our example, we calculate the dot product as \( (1)(2) + (1)(-2) = 2 - 2 = 0 \).

The dot product serves various purposes, such as finding the angle between vectors or determining if two vectors are orthogonal, meaning they are at a 90-degree angle to each other as in this exercise.

Vector Angle

The angle between vectors is a measure that can tell us how vectors are oriented in relation to one another. Finding this angle is crucial in many fields, including physics, where it's necessary for calculating work and torque.

When the dot product of two vectors \( \mathbf{v} \) and \( \mathbf{w} \) is zero, this indicates they are perpendicular, resulting in an angle \( \alpha \) of 90 degrees. In our example, because the dot product came out to be zero, we immediately know that the vectors are orthogonal, simplifying the process of determining the vector angle without further calculation.

Cosine Formula in Precalculus

The Law of Cosines is a powerful tool in precalculus, bridging the gap between right-angled and non-right-angled triangles. This law allows us to calculate the length of any triangle side or the measure of any angle, provided we have sufficient information.

For any triangle with sides \( a, b, \) and \( c \) and the angle \( \alpha \) opposite to side \( a, \) the Law of Cosines states that \( a^2 = b^2 + c^2 - 2bc \cos(\alpha) \). Applied to vectors, it relates vector magnitude, dot product, and the cosine of the vector angle as \( \cos(\alpha) = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| ||\mathbf{w}||} \).

In the example, using the Law of Cosines, we deduced that the cosine of the angle is zero, and hence \( \alpha = 90^\circ \) without the need for a complex calculation. This demonstrates the cosine formula's value in simplifying computations and enhancing our understanding of geometric relationships.