Problem 21 - Precalculus

Question

Find the angle (round to the nearest degree) between each pair of vectors. $$\langle-5 \sqrt{3},-5\rangle \text { and }\langle\sqrt{2},-\sqrt{2}\rangle$$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 79 degrees.

Step 1: Recall the Formula for the Angle Between Two Vectors

The angle \( \theta \) between two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) can be found using the formula: \[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\] where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product of the two vectors, and \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of the vectors \( \mathbf{a} \) and \( \mathbf{b} \) respectively.

Step 2: Calculate the Dot Product

Find the dot product \( \mathbf{a} \cdot \mathbf{b} \) by using the formula: \[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\]For the vectors \( \mathbf{a} = \langle -5\sqrt{3}, -5 \rangle \) and \( \mathbf{b} = \langle \sqrt{2}, -\sqrt{2} \rangle \):\[\mathbf{a} \cdot \mathbf{b} = (-5\sqrt{3})(\sqrt{2}) + (-5)(-\sqrt{2}) = -5\sqrt{6} + 5\sqrt{2}\]

Step 3: Calculate the Magnitudes of the Vectors

The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is calculated by:\[\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2}\]Calculate for \( \mathbf{a} \) and \( \mathbf{b} \):\[\|\mathbf{a}\| = \sqrt{(-5\sqrt{3})^2 + (-5)^2} = \sqrt{75 + 25} = \sqrt{100} = 10\]\[\|\mathbf{b}\| = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2\]

Step 4: Substitute Values Into the Formula for Cosine of the Angle

Substitute the known values into the formula:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \frac{-5\sqrt{6} + 5\sqrt{2}}{10 \times 2} = \frac{-5\sqrt{6} + 5\sqrt{2}}{20}\]

Step 5: Solve for the Angle \( \theta \)

Use the inverse cosine function to find \( \theta \):\[\theta = \cos^{-1}\left(\frac{-5\sqrt{6} + 5\sqrt{2}}{20}\right)\]Calculate \( \theta \) to the nearest degree using a calculator.

Key Concepts

Dot Product

The concept of the dot product is essential when working with vectors. It involves multiplying corresponding components of two vectors and summing these products. For two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), the dot product \(\mathbf{a} \cdot \mathbf{b}\) is calculated as:\[\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2\]

• This formula shows the multiplication of the first components \((a_1 b_1)\). Then, the second components \((a_2 b_2)\) are multiplied together.
• Finally, add these two products together to obtain the dot product value.

Understanding the dot product is crucial because it allows us to compute the angle between two vectors using the cosine formula. It also has numerous applications in physics and engineering. It indicates how much one vector projects onto another.

Vector Magnitude

The magnitude of a vector is like its length. It tells us how big or long the vector is in any dimensional space. When dealing with a vector \(\mathbf{a} = \langle a_1, a_2 \rangle\), the magnitude is calculated by:\[\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2}\]

• This formula starts by squaring each component of the vector. Squaring a number makes it positive, ensuring the total won't be negative—even if the original components were.
• Next, add these squared components together.
• Then, take the square root of the sum to find the magnitude, representing the Euclidean distance from the origin to the point \((a_1, a_2)\).

The magnitude is essential in determining how 'strong' a vector is. Many real-life applications, like determining speed or force, rely on vector magnitude.

Inverse Cosine

To find the angle between two vectors, the inverse cosine function comes into play. Once you've found the cosine of the angle using the formula \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\), you'll need the inverse cosine to find \(\theta\) itself.

• After computing the cosine value, use the inverse cosine function, often written as \(\cos^{-1}\) or \(\text{arccos}\), to find the angle.
• The function \(\cos^{-1}(x)\) tells us what angle \(\theta\) corresponds to a particular cosine value \(x\).
• This function is crucial as it maps a value between -1 and 1 to an angle, usually provided in radians or degrees depending on the context.

Inverse cosine helps translate a computed cosine into a tangible angle measurement, crucial when interpreting the spatial relationship between two vectors.