Question

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

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

use the scalar product to determine the magnitude of vector \\(\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 them.

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

- \(\vec{U}\) is oriented SE (southeast), which is \(45^\circ\) south of east.
- \(\vec{V}\) 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 the south is \(45^\circ\) (since SE is halfway between south and east), and the angle of \(\vec{V}\) from the south is \(15^\circ\) west. So the angle between them \(\theta = 45^\circ - 15^\circ= 30^\circ\)? Wait, no. Wait, let's think in terms of standard position. Wait, maybe better to calculate the angle between their directions. Southeast is \(135^\circ\) from the positive x - axis (if we take east as positive x and north as positive y). \(S\ 15^\circ\ W\) is \(180^\circ + 15^\circ= 195^\circ\) from the positive x - axis? Wait, no. Wait, standard position: 0 degrees is east, 90 degrees is north, 180 degrees is west, 270 degrees is south. So southeast: 45 degrees south of east, so in standard position, that's \(360 - 45= 315^\circ\). \(S\ 15^\circ\ W\): 15 degrees west of south. South is 270 degrees, west of south is towards west, so 270+15 = 285 degrees? Wait, no. Wait, from the south axis, moving 15 degrees towards west. So the angle from the positive x - axis: south is 270 degrees, west of south is adding 15 degrees (since west is clockwise from south), so 270+15 = 285 degrees. The angle of \(\vec{U}\) is 315 degrees, the angle of \(\vec{V}\) is 285 degrees. The angle between them is \(|315 - 285|= 30^\circ\). Wait, that makes sense. So \(\theta = 30^\circ\).

Step3: Use the dot - product formula to solve for \(|\vec{V}|\)

We know that \(\vec{U}\cdot\vec{V}=|\vec{U}||\vec{V}|\cos\theta\). We are given that \(\vec{U}\cdot\vec{V} = 35\), \(|\vec{U}|=7\), and \(\theta = 30^\circ\).
Substitute the known values into the formula:
\(35=7\times|\vec{V}|\times\cos(30^\circ)\)
We know that \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\approx0.866\)
First, we can rewrite the equation as \(|\vec{V}|=\frac{35}{7\times\cos(30^\circ)}\)
Simplify the right - hand side:
\(|\vec{V}|=\frac{5}{\cos(30^\circ)}\)
Since \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\), then \(|\vec{V}|=\frac{5}{\frac{\sqrt{3}}{2}}=\frac{10}{\sqrt{3}}\approx5.77\)? Wait, that can't be right. Wait, maybe I made a mistake in calculating the angle between the vectors.

Wait, let's re - calculate the angle between the vectors. Southeast: direction is \(45^\circ\) south of east. \(S\ 15^\circ\ W\): direction is \(15^\circ\) west of south. So if we take the south direction as a reference, the angle between the two vectors: the vector \(\vec{U}\) is \(45^\circ\) east of south, and the vector \(\vec{V}\) is \(15^\circ\) west of south. So the angle between them is \(45^\circ+ 15^\circ = 60^\circ\). Ah! That's the mistake. So if south is the reference, \(\vec{U}\) is \(45^\circ\) towards east from south, \(\vec{V}\) is \(15^\circ\) towards west from south. So the angle between them is \(45 + 15=60^\circ\). Let's verify with standard position. Southeast: from positive x - axis (east), it's \(360 - 45 = 315^\circ\). \(S\ 15^\circ\ W\): from positive x - axis, south is \(270^\circ\), west of south is \(270+15 = 285^\circ\). The angle between \(315^\circ\) and \(285^\circ\) is \(315 - 285 = 30^\circ\)? Wait, no, the angle between two vectors is the smallest angle between them when placed tail - to - tail. So if we place both vectors tail - to - tail at the origin. \(\vec{U}\) is going to SE (315 degrees), \(\vec{V}\) is going to \(S\ 15^\circ\ W\) (285 degrees). The angle between them is \(315 - 285=30^\circ\)? But when we consider the direction from south, \(\vec{U}\) is 45 degrees east of south, \(\vec{V}\) is 15 degrees west of south. So the angle between them is \(45 + 15=60^\circ\). Let's calculate the angle between the two vectors using their standard position angles.

- Southeast: in standard position (with east as 0 degrees, counter - clockwise), southeast is \(315^\circ\) (because it's 45 degrees below the positive x - axis, so \(360 - 45 = 315\)).
- \(S\ 15^\circ\ W\): south is \(270^\circ\), west of south is moving clockwise from south towards west, so the angle from the positive x - axis is \(270+15 = 285^\circ\) (since moving clockwise from south (270) towards west (180) by 15 degrees, so 270+15 = 285).

The angle between two vectors with angles \(\alpha\) and \(\beta\) from the positive x - axis is \(|\alpha-\beta|\) if we take the smaller angle. \(|315 - 285| = 30^\circ\). Wait, now I'm confused. Let's use a different approach. Let's draw the directions:

- Southeast: the vector is going towards the fourth quadrant, 45 degrees from both south and east.
- \(S\ 15^\circ\ W\): the vector is going towards the third quadrant, 15 degrees west of south (so 15 degrees towards the west from the south direction).

The angle between the south - east vector and the south - west (15 degrees west of south) vector: the south - east vector is 45 degrees east of south, the south - west (15 degrees west of south) vector is 15 degrees west of south. So the angle between them is \(45 + 15=60^\circ\). Because from the south direction, one is 45 degrees to the east, the other is 15 degrees to the west, so the total angle between them is \(45 + 15 = 60^\circ\).

Now, let's re - do the dot - product formula. The dot - product formula is \(\vec{U}\cdot\vec{V}=|\vec{U}||\vec{V}|\cos\theta\), where \(\theta\) is the angle between the two vectors when they are placed tail - to - tail.

We know that \(\vec{U}\cdot\vec{V} = 35\), \(|\vec{U}| = 7\), and we need to find \(|\vec{V}|\). Let's assume that the angle between them is \(60^\circ\) (let's check with both angles).

Case 1: \(\theta = 30^\circ\)
\(35=7\times|\vec{V}|\times\cos(30^\circ)\)
\(|\vec{V}|=\frac{35}{7\times\cos(30^\circ)}=\frac{5}{\frac{\sqrt{3}}{2}}=\frac{10}{\sqrt{3}}\approx5.77\)

Case 2: \(\theta = 60^\circ\)
\(35=7\times|\vec{V}|\times\cos(60^\circ)\)
Since \(\cos(60^\circ)=0.5\)
\(35 = 7\times|\vec{V}|\times0.5\)
\(35=3.5\times|\vec{V}|\)
\(|\vec{V}|=\frac{35}{3.5}=10\)

Now, let's think about the directions again. Southeast is 45 degrees south of east, \(S\ 15^\circ\ W\) is 15 degrees west of south. Let's find the angle between their velocity vectors. The angle between east and west is \(180^\circ\). The vector \(\vec{U}\) is 45 degrees towards south from east, \(\vec{V}\) is 15 degrees towards west from south. So the angle between the east - west components: the east component of \(\vec{U}\) and the west component of \(\vec{V}\). The angle between the two vectors: if we consider the south axis, \(\vec{U}\) is 45 degrees to the east of south, \(\vec{V}\) is 15 degrees to the west of south. So the angle between them is \(45 + 15=60^\circ\). That makes sense. So \(\theta = 60^\circ\).

So using \(\theta = 60^\circ\), \(\cos(60^\circ)=0.5\)

From \(\vec{U}\cdot\vec{V}=|\vec{U}||\vec{V}|\cos\theta\)

\(35 = 7\times|\vec{V}|\times0.5\)

\(35=3.5\times|\vec{V}|\)

\(|\vec{V}|=\frac{35}{3.5}=10\)

Answer:

The magnitude of vector \(\vec{V}\) is \(\boxed{10}\) units.