Vector Cross Product Calculation Explained

This problem involves finding the magnitude of the cross product of two vectors, \(\vec a\) and \(\vec b\), given their individual magnitudes and their dot product. We will use fundamental vector formulas and trigonometric identities to solve this problem step-by-step.

Vector Magnitudes and Dot Product

We are provided with the following information about the vectors \(\vec a\) and \(\vec b\):

• The magnitude of vector \(\vec a\), denoted as \(\left| {\vec a\;} \right|\), is 13.
• The magnitude of vector \(\vec b\), denoted as \(\left| {\vec b} \right|\), is 5.
• The dot product of vector \(\vec a\) and vector \(\vec b\), denoted as \(\vec a\;.\;\vec b\), is 60.

Our objective is to determine the value of the magnitude of the cross product, \(\left| {\vec a \times \vec b} \right|\).

Cosine Calculation from Dot Product

The dot product of two vectors \(\vec a\) and \(\vec b\) is defined by the formula:

\[ \vec a\;.\;\vec b = \left| {\vec a\;} \right| \left| {\vec b} \right| \cos \theta \]

where \(\theta\) represents the angle between the vectors \(\vec a\) and \(\vec b\). We can substitute the given values into this formula:

\[ 60 = (13)(5) \cos \theta \]

First, calculate the product of the magnitudes:

\[ 60 = 65 \cos \theta \]

Now, to find \(\cos \theta\), divide both sides by 65:

\[ \cos \theta = \frac{60}{65} \]

Simplifying the fraction by dividing both the numerator and the denominator by 5, we get:

\[ \cos \theta = \frac{12}{13} \]

Sine Determination using Trigonometric Identity

To calculate the magnitude of the cross product, we need the value of \(\sin \theta\). We can find this using the fundamental trigonometric identity relating sine and cosine:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Rearranging the formula to solve for \(\sin^2 \theta\):

\[ \sin^2 \theta = 1 - \cos^2 \theta \]

Now, substitute the value of \(\cos \theta\) we just found:

\[ \sin^2 \theta = 1 - \left( \frac{12}{13} \right)^2 \]

Calculate the square of \(\frac{12}{13}\):

\[ \sin^2 \theta = 1 - \frac{144}{169} \]

To subtract, convert 1 to a fraction with the same denominator as \(\frac{144}{169}\):

\[ \sin^2 \theta = \frac{169}{169} - \frac{144}{169} \]

Perform the subtraction:

\[ \sin^2 \theta = \frac{169 - 144}{169} \]

\[ \sin^2 \theta = \frac{25}{169} \]

Finally, take the square root of both sides to find \(\sin \theta\). Since the angle \(\theta\) between two vectors is conventionally taken in the range \(0 \le \theta \le \pi\), \(\sin \theta\) will be non-negative:

\[ \sin \theta = \sqrt{\frac{25}{169}} \]

\[ \sin \theta = \frac{5}{13} \]

Cross Product Magnitude Calculation

The magnitude of the cross product of two vectors \(\vec a\) and \(\vec b\) is given by the formula:

\[ \left| {\vec a \times \vec b} \right| = \left| {\vec a\;} \right| \left| {\vec b} \right| \sin \theta \]

Now, substitute the given magnitudes of the vectors and the calculated value of \(\sin \theta\) into this formula:

\[ \left| {\vec a \times \vec b} \right| = (13)(5) \left( \frac{5}{13} \right) \]

First, multiply the magnitudes of the vectors:

\[ \left| {\vec a \times \vec b} \right| = 65 \times \frac{5}{13} \]

We can simplify this by noticing that 13 is a factor of 65 (\(65 = 5 \times 13\)):

\[ \left| {\vec a \times \vec b} \right| = (5 \times 13) \times \frac{5}{13} \]

Cancel out the 13 in the numerator and denominator:

\[ \left| {\vec a \times \vec b} \right| = 5 \times 5 \]

\[ \left| {\vec a \times \vec b} \right| = 25 \]

Thus, the value of \(\left| {\vec a \times \vec b} \right|\) is 25.

Final Answer Summary

By using the definition of the dot product to find the cosine of the angle between the vectors, then employing a trigonometric identity to find the sine of the angle, and finally applying the formula for the magnitude of the cross product, we arrived at the solution.

The calculated value of \(\left| {\vec a \times \vec b} \right|\) is 25.