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 \(69.3^\circ\).