Question

vector a has magnitude 13.0 m and vector b has magnitude 15.0 m. the scalar product \\(\vec{a} \cdot \vec{b}\\) is 105 m². what is the magnitude of the vector product between these two vectors? express your answer with the appropriate units.

Explanation:

Step1: Recall scalar and vector product formulas

The scalar product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta\), where \(|\vec{A}|\) and \(|\vec{B}|\) are the magnitudes of the vectors, and \(\theta\) is the angle between them. The magnitude of the vector product (cross product) is given by \(|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin\theta\).

We know \(|\vec{A}| = 13.0\space m\), \(|\vec{B}| = 15.0\space m\), and \(\vec{A} \cdot \vec{B} = 105\space m^2\). First, we can find \(\cos\theta\) from the scalar product formula:

\(\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}\)

Substituting the given values:

\(\cos\theta = \frac{105}{13.0 \times 15.0}\)

Step2: Calculate \(\cos\theta\)

First, calculate the denominator: \(13.0 \times 15.0 = 195\)

Then, \(\cos\theta = \frac{105}{195} = \frac{7}{13} \approx 0.5385\)

Step3: Find \(\sin\theta\)

We know that \(\sin^2\theta + \cos^2\theta = 1\), so \(\sin\theta = \sqrt{1 - \cos^2\theta}\)

Substituting \(\cos\theta \approx 0.5385\):

\(\sin\theta = \sqrt{1 - (0.5385)^2} = \sqrt{1 - 0.290} = \sqrt{0.710} \approx 0.8426\)

Step4: Calculate the magnitude of the vector product

Using the formula \(|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin\theta\)

Substitute \(|\vec{A}| = 13.0\space m\), \(|\vec{B}| = 15.0\space m\), and \(\sin\theta \approx 0.8426\):

\(|\vec{A} \times \vec{B}| = 13.0 \times 15.0 \times 0.8426\)

First, calculate \(13.0 \times 15.0 = 195\)

Then, \(195 \times 0.8426 \approx 164.3\)

Answer:

The magnitude of the vector product is approximately \(\boxed{164}\space m^2\) (rounded to a reasonable number of significant figures).