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The dot product, also known as the scalar product, is a way of combining two vectors to result in a single number, or scalar. It provides insights into how much of one vector goes in the direction of another. To find the dot product of two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), use the formula:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]In our exercise, vectors \( \mathbf{a} = (1, -1) \) and \( \mathbf{b} = (3, 1) \) give a dot product of:

• \( 1 \times 3 + (-1) \times 1 = 3 - 1 = 2 \)

The dot product can be thought of as a projection, where you project one vector onto another. If the dot product is positive, as it is here, the vectors are pointing somewhat in the same general direction.

The magnitude of a vector, often referred to as its length or norm, tells us how long the vector is. Imagine you're measuring the length of an arrow in space—the magnitude is this measure. For a vector \( \mathbf{a} = (a_1, a_2) \), the magnitude \( |\mathbf{a}| \) is found using the Pythagorean theorem:\[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2} \]For \( \mathbf{a} = (1, -1) \):

• \( |\mathbf{a}| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \)

For \( \mathbf{b} = (3, 1) \):

• \( |\mathbf{b}| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \)

Understanding vector magnitude is crucial in applications ranging from physics to computer graphics, where it helps in normalizing vectors and calculating distances.

The inverse cosine function, denoted as \( \cos^{-1} \) or \( \arccos \), helps us find the angle between two vectors when we know their dot product and magnitudes. When vectors \( \mathbf{a} \) and \( \mathbf{b} \) have a known dot product \( \mathbf{a} \cdot \mathbf{b} \) and magnitudes \( |\mathbf{a}| \) and \( |\mathbf{b}| \), the angle \( \theta \) is found using the formula:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot |\mathbf{b}|} \]Thus, to find \( \theta \), we need to use the inverse cosine function:

• \( \theta = \cos^{-1} \left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot |\mathbf{b}|} \right) \)
• Here, \( \theta = \cos^{-1} \left( \frac{2}{\sqrt{2} \times \sqrt{10}} \right) \approx 63.43^\circ \)

Using a calculator is the simplest way to compute this, and it helps translate the abstract concept into a concrete angle value, showing how these vectors orient in space relative to each other.