Question

If \(\mathrm{A}^{\rightarrow}=4 \mathrm{i} \wedge+3 \mathrm{j} \wedge-2 \mathrm{k} \wedge\) and \(\mathrm{B}^{\rightarrow}=8 \mathrm{i} \wedge+6 \mathrm{j} \wedge-4 \mathrm{k} \wedge\) the angle between \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) is (A) \(45^{\circ}\) (B) \(0^{\circ}\) (C) \(60^{\circ}\) (D) \(90^{\circ}\)

Short answer

The angle between vectors A and B is \( 0^{\circ} \) (Option B).

Step-by-step solution

Define the given vectors

The given vectors are: \( A = 4i + 3j - 2k \) and \( B = 8i + 6j - 4k \)

Find the dot product of A and B

The dot product of two vectors A and B is given by the formula: \( A \cdot B = |A|\,|B| \, \cos{\theta} \) Where |A| and |B| are the magnitudes of vectors A and B, respectively, and θ is the angle between A and B. Now, let's find the dot product: \( A \cdot B = (4)(8) + (3)(6) + (-2)(-4) = 32 + 18 + 8 = 58 \)

Find the magnitudes of A and B

Next, let's find the magnitudes of A and B: magnitude of A, \( |A| = \sqrt{ 4^2 + 3^2 + (-2)^2} = \sqrt{29} \) magnitude of B, \( |B| = \sqrt{ 8^2 + 6^2 + (-4)^2} = 2\sqrt{29} \)

Plug magnitude values into the formula

Now we have everything required to plug the values into the formula: \( 58 = (\sqrt{29})(2\sqrt{29}) \, \cos{\theta} \)

Solve for the angle θ

By solving for θ, we get: \( \cos{\theta} = \frac{58}{58} = 1 \) Using the inverse cosine function, we find the angle θ: \( \theta = \cos^{-1}(1) \) \( \theta = 0^{\circ} \) So, the angle between vectors A and B is \( 0^{\circ} \)（Option B）.

Key concepts

Vector Magnitude

When dealing with vectors, one of the primary steps is understanding their magnitude. Think of vector magnitude as the "length" of the vector in the geometric sense. It tells us how "long" or "strong" a vector is. For a vector given in three-dimensional space, such as \(A = 4i + 3j - 2k\), the magnitude \(|A|\) is calculated using the formula: \(\sqrt{a^2 + b^2 + c^2}\), where \(a\), \(b\), and \(c\) are the components of the vector along the \(i\), \(j\), and \(k\) axes, respectively.
In the exercise, the magnitude of vector \(A\) is computed as follows:

• Square each component: \(4^2 = 16\), \(3^2 = 9\), \((-2)^2 = 4\).
• Add these squares: \(16 + 9 + 4 = 29\).
• Take the square root of the sum: \(\sqrt{29}\).

Repeat these steps for any vector to find its magnitude. The magnitude gives you a scalar quantity, which is crucial for further calculations, like the dot product or projections.

Cosine of Angle

The cosine of the angle between two vectors provides valuable insight into their relative orientation. Mathematically, the dot product of two vectors \(A\) and \(B\) is given by \(A \cdot B = |A||B|\cos{\theta}\), where \(|A|\) and \(|B|\) are the magnitudes of \(A\) and \(B\), respectively, and \(\theta\) denotes the angle between them. The cosine value helps determine:

• If the vectors are parallel, \(\cos{\theta} = 1\).
• If the vectors are perpendicular, \(\cos{\theta} = 0\).
• If the vectors point in opposite directions, \(\cos{\theta} = -1\).

In our problem, after computing the magnitudes and the dot product, we solved for \(\cos{\theta}\) and found it to be 1. This tells us that the vectors \(A\) and \(B\) are in the same direction or parallel.

Angle Between Vectors

Determining the angle between two vectors is crucial in understanding their spatial relationship. This angle, \(\theta\), can be found using the dot product formula: \(A \cdot B = |A||B|\cos{\theta}\). In simpler terms, if you have the dot product and the magnitudes of the vectors, you can easily find the angle, which shows us:

• \(\theta\) close to \(0^{\circ}\) implies vectors point in nearly the same direction.
• \(\theta\) close to \(90^{\circ}\) implies vectors are perpendicular.
• \(\theta\) close to \(180^{\circ}\) implies vectors point in opposite directions.

In the exercise, after calculating \(\cos{\theta}\) as 1, we determined that \(\theta = 0^{\circ}\). This means vector \(A\) is completely aligned with vector \(B\), confirming they are essentially the same vector scaled differently, which results in an angle of \(0^{\circ}\) between them. Understanding this helps visualize how vector orientation affects calculations and interpretations in physics and engineering situations.