Question

Two vectors have moduli 7 and 13 respectively. The angle between them is \(45^{\circ}\). Evaluate their scalar product.

Short answer

Answer: The scalar product of the two given vectors is \(A \cdot B = 91 \cdot \frac{\sqrt{2}}{2}\).

Step-by-step solution

Convert degrees to radians

Since trigonometric functions often deal with radians, convert the angle from degrees to radians: \(\theta = 45^{\circ} \cdot \frac{\pi}{180} = \frac{\pi}{4}\)

Write down the magnitudes and angle

Now, we have the magnitudes of both vectors and the angle between them: |A| = 7, |B| = 13 and \(\theta = \frac{\pi}{4}\)

Use the scalar product formula

Using the scalar product formula, we can calculate the scalar product of the two vectors: \(A \cdot B = |A||B|\cos\theta\)

Insert the given values into the formula

Put the given values of magnitudes and angle into the formula: \(A \cdot B = 7 \cdot 13 \cdot \cos\frac{\pi}{4}\)

Compute the scalar product

Now, calculate the scalar product using the given values: \(A \cdot B = 7 \cdot 13 \cdot \frac{\sqrt{2}}{2} = 91 \cdot \frac{\sqrt{2}}{2}\) So, the scalar product of the two given vectors is: \(A \cdot B = 91 \cdot \frac{\sqrt{2}}{2}\).

Key concepts

Scalar Product

The scalar product, also known as the dot product, is a fundamental concept in vector calculus. It allows us to find the product of two vectors in terms of magnitude and direction. Defined for two vectors \( \mathbf{A} \) and \( \mathbf{B} \), the scalar product is given by:

• \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \)

The scalar product essentially combines the magnitudes of the vectors with the cosine of the angle \( \theta \) between them. This captures both the size and directional alignment of the vectors. It results in a scalar quantity—not a vector—hence its name. This concept is particularly useful when analyzing projections and determining if two vectors are perpendicular (when the dot product is zero).
To use this formula, just multiply the magnitudes of the vectors by the cosine of the angle between them.

Trigonometric Conversion

Many times, angles are provided in degrees, but calculus and physics problems use radians for calculations. That's where trigonometric conversion comes into play.

• Converting degrees to radians is necessary, because the trigonometric functions of most mathematical formulas operate in radian mode.

The conversion is simple:

• Use the formula: \( \theta_{radians} = \theta_{degrees} \times \frac{\pi}{180} \)

For our example problem, converting an angle of \( 45^{\circ} \), you calculate:

• \( 45^{\circ} \times \frac{\pi}{180} = \frac{\pi}{4} \)

This allows you to accurately incorporate trigonometric functions like cosine into the scalar product formula without error.

Calculating Magnitudes

Before computing with vectors, understanding their magnitudes is crucial. The magnitude (or length) of a vector \( \mathbf{A} \) is shown by \( |\mathbf{A}| \), and it's computed using the square root of the sum of the square of its components if they are known.

• For example, a vector \( \mathbf{A} = (a, b, c) \) has magnitude: \( |\mathbf{A}| = \sqrt{a^2 + b^2 + c^2} \)

In our exercise, the magnitudes of vectors \( \mathbf{A} \) and \( \mathbf{B} \) are directly given, not computed from components:

• |\( A\)| = 7
• |\( B\)| = 13

Understanding this helps apply the scalar product formula as a straightforward multiplication of these magnitudes with the cosine of the angle between the vectors.

Angle Between Vectors

The angle between two vectors is another important aspect of vector calculus, critical when computing the scalar product. This angle is denoted by \( \theta \), and it determines how aligned the vectors are.

• When \( \theta = 0 \) degrees, vectors are perfectly aligned, which means they point in the same direction.
• When \( \theta = 90^{\circ} \) or \( \frac{\pi}{2} \) radians, vectors are perpendicular, resulting in a scalar product of zero.

The cosine term in the scalar product formula directly depends on this angle, affecting the final result. For \( \theta = 45^{\circ} \) or \( \frac{\pi}{4} \) radians, used in our example, the cosine value shrinks the product result due to partial alignment.

• The trigonometric identity used here is \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).

Understanding how angle impacts the dot product enables better intuition for vector projection and directional behavior.