Problem 72 - A Graphical Approach to Precalculus with Limits

Question

Find the angle between each pair of vectors. $$3 \mathfrak{i}+4 \mathfrak{j}, \mathfrak{j}$$

Step-by-Step Solution

Answer

The angle between the vectors is \(\cos^{-1}(\frac{4}{5})\).

Step 1: Understand the Problem

We are given two vectors: the first vector \( \mathbf{A} = 3\mathfrak{i} + 4\mathfrak{j} \) and the second vector \( \mathbf{B} = \mathfrak{j} \). We need to find the angle between these vectors using the dot product formula.

Step 2: Dot Product Formula

The formula for the dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by: \[ \mathbf{A} \cdot \mathbf{B} = \|\mathbf{A}\|\|\mathbf{B}\|\cos\theta \] where \( \theta \) is the angle between the two vectors. Rearrange this to find: \[ \cos\theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\|\|\mathbf{B}\|} \]

Step 3: Compute the Dot Product

Calculate the dot product of \( \mathbf{A} = 3\mathfrak{i} + 4\mathfrak{j} \) and \( \mathbf{B} = \mathfrak{j} \). \[ \mathbf{A} \cdot \mathbf{B} = (3)(0) + (4)(1) = 4 \]

Step 4: Find the Magnitude of Each Vector

Calculate the magnitude of each vector. For \( \mathbf{A} \): \[ \|\mathbf{A}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]For \( \mathbf{B} \): \[ \|\mathbf{B}\| = \sqrt{0^2 + 1^2} = 1 \]

Step 5: Calculate Cosine of the Angle

Substitute the dot product and magnitudes into the cosine formula:\[ \cos\theta = \frac{4}{5 \times 1} = \frac{4}{5} \]

Step 6: Calculate the Angle \(\theta\)

Find the angle \(\theta\) by taking the inverse cosine:\[ \theta = \cos^{-1}\left(\frac{4}{5}\right) \]. This angle, \(\theta\), is the one we need.

Key Concepts

Dot Product

The dot product of two vectors is an algebraic operation that combines the components of two vectors. Imagine you have two vectors, \( \mathbf{A} = 3\mathfrak{i} + 4\mathfrak{j} \) and \( \mathbf{B} = \mathfrak{j} \). You can calculate the dot product by following the formula:

• Multiply the corresponding components of the vectors.
• For vectors \( \mathbf{A} \) and \( \mathbf{B} \), this means multiplying the \( \mathfrak{i} \) components together and the \( \mathfrak{j} \) components together.
• Add the results of these multiplications.

For our example, it will be \( 3 \cdot 0 + 4 \cdot 1 = 4 \). The dot product, therefore, results in a scalar value, not a vector.
The importance of the dot product in vector calculus can't be overstated. It's crucial for defining the angle between vectors and for determining orthogonality. A dot product of zero indicates vectors are perpendicular.

Magnitude of a Vector

The magnitude of a vector, sometimes referred to as the vector's length or norm, tells us how long the vector is. It's calculated using the Pythagorean theorem.

• Take each component of the vector, square it, and then sum these squares.
• Finally, take the square root of this sum.

For the vector \( \mathbf{A} = 3\mathfrak{i} + 4\mathfrak{j} \), the magnitude \( \|\mathbf{A}\| \) is \( \sqrt{3^2 + 4^2} \) which equals to \( \sqrt{9 + 16} = 5 \).
Meanwhile, for \( \mathbf{B} = \mathfrak{j} \), since there is only a \( \mathfrak{j} \) component, it's \( \|\mathbf{B}\| = 1 \). Knowing the magnitude is key for understanding both the vector's direction and force.

Angle Between Vectors

To find the angle between two vectors, use the dot product formula combined with their magnitudes.

• The formula is \( \cos\theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\| \|\mathbf{B}\|} \).
• You need the dot product and each vector's magnitude.
• Calculate the arccosine (inverse cosine) of the result to find the angle.

For vectors \( \mathbf{A} = 3\mathfrak{i} + 4\mathfrak{j} \) and \( \mathbf{B} = \mathfrak{j} \), the prior steps yield \( \cos\theta = \frac{4}{5} \). Finally, \( \theta = \cos^{-1}(\frac{4}{5}) \).
This angle tells us how directions are aligned. Smaller angles mean vectors point similarly, while larger ones mean they diverge.