Question

question 13 of 16
consider the information given below about vectors \\(\boldsymbol{\vec{u}}\\) and \\(\boldsymbol{\vec{v}}\\).

• \\(\boldsymbol{\vert \vec{u} \vert = 7}\\) units
• \\(\boldsymbol{\vec{u} \bullet \vec{v} = 35}\\)
• \\(\boldsymbol{\vec{u}}\\) is oriented se
• \\(\boldsymbol{\vec{v}}\\) is oriented s \\(15^\circ\\) w

use the scalar product to determine the magnitude of vector \\(\boldsymbol{\vec{v}}\\).
please round your answer to the nearest unit.

Explanation:

Step1: Recall the scalar product formula

The scalar product (dot product) of two vectors \(\vec{U}\) and \(\vec{V}\) is given by \(\vec{U} \cdot \vec{V} = |\vec{U}| |\vec{V}| \cos\theta\), where \(\theta\) is the angle between the two vectors.

Step2: Determine the angle between \(\vec{U}\) and \(\vec{V}\)

• \(\vec{U}\) is oriented SE (southeast), which means it makes a \(45^\circ\) angle with the south (or east) direction.
• \(\vec{V}\) is oriented \(S 15^\circ W\) (15 degrees west of south).
• To find the angle between them, we calculate the difference in their angles from the south direction. The angle of \(\vec{U}\) from south is \(45^\circ\) (towards east), and the angle of \(\vec{V}\) from south is \(15^\circ\) (towards west). So the angle between them \(\theta = 45^\circ + 15^\circ = 60^\circ\).

Step3: Solve for \(|\vec{V}|\)

We know \(\vec{U} \cdot \vec{V} = 35\), \(|\vec{U}| = 7\), and \(\theta = 60^\circ\). From the dot product formula:
\[
35 = 7 \times |\vec{V}| \times \cos(60^\circ)
\]
Since \(\cos(60^\circ) = 0.5\), we substitute that in:
\[
35 = 7 \times |\vec{V}| \times 0.5
\]
Simplify the right - hand side: \(7\times0.5 = 3.5\), so the equation becomes:
\[
35 = 3.5\times|\vec{V}|
\]
To solve for \(|\vec{V}|\), divide both sides by \(3.5\):
\[
|\vec{V}|=\frac{35}{3.5}=10
\]

Answer:

10