Problem 80 - 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 \(0^{\circ} \leq \alpha \leq 180^{\circ} . )\) $$\mathbf{v}=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{w}=2 \mathbf{i}-\mathbf{j}$$

Step-by-Step Solution

Answer

\(\alpha = 90^{\circ}\).

Step 1: Calculation of the lengths of vectors

Calculate the lengths of the vectors \(v\) and \(w\). The length of a vector \(a = a_i + a_j\), is calculated as \(|a| = \sqrt{a_i^2 + a_j^2}\). In this problem, \(v = i + 2j\), so \(|v| = \sqrt{1^2 + 2^2} = \sqrt{5}\). Similarly, for \(w = 2i -j\), \(|w| = \sqrt{2^2 + (-1)^2} = \sqrt{5}\).

Step 2: Calculation of dot product

The dot product \(v.w\) of two vectors \(v = v_i + v_j\) and \(w = w_i + w_j\) is given by \(v.w = v_i*w_i + v_j*w_j\). For \(v = i + 2j\) and \(w = 2i -j\), \(v.w = 1*2 + 2*(-1) = 0\).

Step 3: Application of the Law of Cosines for vectors

The angle \(\alpha\) between vectors v and w can be calculated using the law of cosines formulated for vectors which is given by \(\cos(\alpha) = \frac{v.w}{|v|*|w|}\). Substituting the calculated values, we have \(\cos(\alpha) = \frac{0}{\sqrt{5}*\sqrt{5}} = 0\). Therefore, \(\alpha = \arccos(0)\).

Step 4: Conversion to degrees

The arccos of 0 is 90 degrees in the interval [0°,180°]. Hence, \(\alpha = 90^{\circ}\).

Key Concepts

Understanding the Vector Angle

The angle between two vectors tells us how much one vector needs to be rotated to align with another vector.
It's an important aspect in various applications like physics and engineering.
The vector angle is typically found using the dot product and vector magnitudes together in an equation. To understand the angle between vectors, think of how pulling a cart at different angles requires different amounts of effort.
If vectors represent forces or displacements, knowing the angle helps analyze such real-world problems.

Exploring the Dot Product

The dot product (also known as the scalar product) is an operation that takes two vectors and generates a scalar.
It's like multiplying two numbers but in a way that's specific to vectors.
For two 2D vectors, \ \(\mathbf{v} = v_i\mathbf{i} + v_j\mathbf{j}\) and \ \(\mathbf{w} = w_i\mathbf{i} + w_j\mathbf{j}\), the dot product is calculated as \ \(v \cdot w = v_i \cdot w_i + v_j \cdot w_j\).
This scalar result can tell us a lot about the relationship between vectors:

• If the dot product is positive, vectors point in a generally similar direction.
• If the dot product is zero, vectors are perpendicular.
• If the dot product is negative, vectors point in opposite directions.

Therefore, the dot product is fundamental in calculating the angle between two vectors using \\( \cos(\alpha) = \frac{v \cdot w}{|v| |w|} \).

Calculating Vector Magnitude

The magnitude of a vector, often called the length, shows how long the vector is.
You can think of it like the hypotenuse in a right-angled triangle.
To calculate a vector's magnitude in 2D, like \\( \mathbf{v} = v_i\mathbf{i} + v_j\mathbf{j} \), we use the formula: \ \(|\mathbf{v}| = \sqrt{v_i^2 + v_j^2}\).
This formula comes from the Pythagorean Theorem, explaining that each vector component creates a right triangle.
Magnitude is always a positive number, showcasing only the size of the vector without directional information.
In the vector angle problem, vector magnitude helps scale the dot product answer to ultimately find the angle.

Trigonometric Identity: Arccosine

Trigonometric identities link angles and side lengths in triangles and are crucial in many math problems.
One important identity is the arccosine function, also written as \\(\arccos\).
It helps find an angle when the cosine of that angle is known. If \\(\cos(\alpha)\) is the cosine of an angle, then \\(\alpha = \arccos(\cos(\alpha))\).
This inverse function essentially "undoes" the cosine, providing the original angle.
For our vector problem, \\(\alpha = \arccos(0)\) indicates that the angle between the vectors is 90 degrees, which is easily interpreted in standard math applications.
Understanding this identity helps in solving many geometric and trigonometric problems.