Question

Find the acute angle between the two vectors \(\mathbf{A}=2 \mathbf{a}_{x}+\mathbf{a}_{y}+3 \mathbf{a}_{z}\) and \(\mathbf{B}=\mathbf{a}_{x}-3 \mathbf{a}_{y}+2 \mathbf{a}_{z}\) by using the definition of \((a)\) the dot product; \((b)\) the cross product.

Short answer

Answer: The acute angle between the two vectors is approximately \(1.21\) radians or \(69.3^\circ\).

Step-by-step solution

(a) Using the Dot Product

To find the angle between the two vectors using the dot product, we'll follow these steps: 1. Calculate the dot product of the two vectors (\(\mathbf{A} \cdot \mathbf{B}\)). 2. Calculate the magnitudes of both vectors (\(|\mathbf{A}|\) and \(|\mathbf{B}|\)). 3. Use these values and the dot product formula to find the cosine of the angle between the vectors. The formula is \(\cos\theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\) 4. Calculate the angle in radians and degrees.

Calculate the Dot Product

The dot product of vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by: \(\mathbf{A} \cdot \mathbf{B} = (2)(1) + (1)(-3) + (3)(2) = 2 - 3 + 6 = 5\).

Calculate the Magnitudes

The magnitudes of vectors \(\mathbf{A}\) and \(\mathbf{B}\) are given by: \(|\mathbf{A}| = \sqrt{(2)^2 + (1)^2 + (3)^2} = \sqrt{14}\) and \(|\mathbf{B}| = \sqrt{(1)^2 + (-3)^2 + (2)^2} = \sqrt{14}\).

Calculate the Cosine of the Angle

Using the formula for the cosine of the angle between the two vectors, we have: $$\cos(\theta) = \frac{5}{\sqrt{14} \sqrt{14}} = \frac{5}{14}$$

Calculate the Angle in Radians and Degrees

To find the angle, we take the inverse cosine (also known as arccos) of the value found above: $$\theta = \arccos \left(\frac{5}{14}\right) \approx 1.21 \text{ radians}$$ To convert to degrees, use the formula: $$\theta = \frac{180}{\pi} \times 1.21 \approx 69.3^\circ$$.

(b) Using the Cross Product

To find the angle between the two vectors using the cross product, we'll follow these steps: 1. Calculate the cross product of the two vectors (\(\mathbf{A} \times \mathbf{B}\)). 2. Calculate the magnitudes of the cross product vector (\(|\mathbf{A} \times \mathbf{B}|\)). 3. Use these values and the cross product formula to find the sine of the angle between the vectors. The formula is \(\sin\theta = \frac{|\mathbf{A} \times \mathbf{B}|}{|\mathbf{A}||\mathbf{B}|}\) 4. Calculate the angle in radians and degrees using this sine value.

Calculate the Cross Product

The cross product of vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by: $$\mathbf{A} \times \mathbf{B} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} (1)(2)-(-3)(3) \\ (3)(1)-(2)(2) \\ (2)(-3)-(1)(1) \end{pmatrix} = \begin{pmatrix} 11 \\ -1 \\ -7 \end{pmatrix} $$

Calculate the Magnitude of the Cross Product Vector

The magnitude of the cross product vector is given by: $$|\mathbf{A} \times \mathbf{B}| = \sqrt{(11)^2 + (-1)^2 + (-7)^2} = \sqrt{141}$$

Calculate the Sine of the Angle

Using the formula for the sine of the angle between the two vectors, we have: $$\sin(\theta) = \frac{\sqrt{141}}{\sqrt{14} \sqrt{14}} = \frac{\sqrt{141}}{14} $$

Calculate the Angle in Radians and Degrees

To find the angle, we take the inverse sine (also known as arcsin) of the value found above: $$\theta' = \arcsin \left(\frac{\sqrt{141}}{14}\right) \approx 1.21 \text{ radians}$$ To convert to degrees, use the formula: $$\theta' = \frac{180}{\pi} \times 1.21 \approx 69.3^\circ$$. Both methods yield the same result: the acute angle between the two vectors is approximately \(1.21\) radians or \(69.3^\circ\).

Key concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental concept in vector algebra. It's an easy way to determine the relationship between two vectors as it gives insight into the angle between them. The dot product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) is defined as the sum of the products of their corresponding components:

• Formula: \( \mathbf{A} \cdot \mathbf{B} = A_1B_1 + A_2B_2 + A_3B_3 \).
• In this specific exercise, the dot product was calculated as \( 5 \).

A positive dot product indicates that the vectors are pointing in a generally similar direction, while a negative one means they point in opposite directions. If the dot product is zero, the vectors are perpendicular. The formula involving the cosine of the angle between the vectors is given by:

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

This relationship helps us find the angle between the two vectors.

Cross Product

The cross product, unlike the dot product, results in another vector instead of a scalar. It's crucial for finding a vector that is perpendicular to the initial two vectors. The cross product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) is calculated by:

• Formula: \( \mathbf{A} \times \mathbf{B} = (A_2B_3 - A_3B_2, A_3B_1 - A_1B_3, A_1B_2 - A_2B_1) \).
• In the given exercise: \( \mathbf{A} \times \mathbf{B} = (11, -1, -7) \).

The magnitude of the resulting vector indicates the parallelogram area spanned by the two original vectors. When using the cross product to find the angle between the vectors, we use the sine of the angle:

• \( \sin(\theta) = \frac{|\mathbf{A} \times \mathbf{B}|}{|\mathbf{A}||\mathbf{B}|} \).

This equation is especially useful when assessing the relative orientation of vectors.

Angle Between Vectors

Determining the angle between vectors is an important task in vector algebra. Whether you are using the dot or cross product, the angle between two vectors \( \theta \) can reveal much about their spatial relationship:

• Acute angles indicate vectors pointing more towards each other.
• Obtuse angles indicate vectors diverging away.

In the exercise, both methods calculated the angle between vectors \( \mathbf{A} \) and \( \mathbf{B} \) to be approximately \( 69.3^\circ \). This was achieved either by taking the inverse cosine or inverse sine of their respective values calculated via dot or cross product methods. Recognizing whether vectors form an acute or obtuse angle can help in understanding their interaction in physical spaces.

Vector Magnitudes

Vector magnitudes give us an idea of the length or size of a vector in space. They are an essential part of vector operations, especially when finding angles or performing normalization:

• The magnitude of a vector \( \mathbf{A} = (A_1, A_2, A_3) \) is calculated using: \( |\mathbf{A}| = \sqrt{A_1^2 + A_2^2 + A_3^2} \).
• For the vectors in the exercise, both \( \mathbf{A} \) and \( \mathbf{B} \) have magnitudes of \( \sqrt{14} \).

Understanding vector magnitudes is crucial as it relates to both the length of the vectors and as input to other calculations such as the dot or cross product. The larger the magnitude, the longer the vector, which can significantly influence geometric interpretations in multidimensional spaces.