# How can I find the resultant of a set of vectors which are inside of a circle?

#### Chemist116

The problem is as follows:

Find the resultant vector from the figure from below: $\begin{array}{ll} 1.&10\\ 2.&20\\ 3.&30\\ 4.&10\sqrt{7}\\ 5.&20\sqrt{7}\\ \end{array}$

What I've attempted to do was to find the resultant as follows:

$R=\vec{v}+\vec{w}+\vec{u}+\vec{c}+\vec{u}+\vec{b}$

$\vec{v}+\vec{w}+\vec{u}=2\vec{c}$

$2\vec{c}+\vec{u}=\vec{b}$

Then

$R=2\vec{c}+\vec{c}+\vec{u}+2\vec{c}+\vec{u}$

$R=5\vec{c}+2\vec{u}=5(10)+2(\sqrt{3})$

However, this answer $50+2\sqrt{3}$ doesn't appear in any of the alternatives. What could it be wrong? Help! Please.

#### skeeter

Math Team
$|2\vec{b} + \vec{c}| = 20\sqrt{7}$

#### skipjack

Forum Staff
$R=\vec{v}+\vec{w}+\vec{u}+\vec{c}+\vec{u}+\vec{b}$
Shouldn't the second $\vec{u}$ be $\vec{a}$?

#### Chemist116

Shouldn't the second u⃗ u→ be a⃗ a→?
Sorry that was a mistake from my end. What I intended to say was:

$R=\vec{v}+\vec{w}+\vec{u}+\vec{a}+\vec{b}+\vec{c}$

Then:

$R=2\vec{c}+\vec{a}+\vec{b}+\vec{c}$

$R=3\vec{c}+\vec{a}+\vec{b}$

But:

$2\vec{c}+\vec{a}=\vec{b}$

Then:

$R=3\vec{c}+\vec{a}+2\vec{c}+\vec{a}=5\vec{c}+2\vec{a}$

$\vec{c}=10$

$\vec{a}=10\sqrt{3}$

Therefore:

$50+20\sqrt{3}$

Where and why in this part is the error?.

#### Chemist116

$|2\vec{b} + \vec{c}| = 20\sqrt{7}$
How did you got to that result?

I ended up with:

$\vec{R}=\vec{v}+\vec{w}+\vec{u}+\vec{a}+\vec{b}+\vec{c}$

$\vec{R}=2\vec{c}+\vec{a}+\vec{b}+\vec{c}$

$\vec{R}=3\vec{c}+\vec{a}+\vec{b}$

$2\vec{c}+\vec{a}=\vec{b}$

$\vec{R}=3\vec{c}+\vec{a}+2\vec{c}+\vec{a}$

$\vec{R}=5\vec{c}+2\vec{a}$

Then this is:

$\vec{c}=50$

$\vec{a}=20\sqrt{3}$

Then using the cosines law:

$\left\|5\vec{c}+2\vec{a}\right\|^2=50^2+(20\sqrt{3})^2-2(50)(20\sqrt{3})\cos 30^{\circ}$

$\left\|5\vec{c}+2\vec{a}\right\|=10\sqrt{7}$

@skeeter Am I right? Or could it be that I did some misinterpretation? Help please?

#### skipjack

Forum Staff
You are correct (skeeter slipped up).

Math Team

#### pto189

Your error is 2c+u =/ b.
We have v+w+u+a+b+c = b+b+c = 2b+c
Using Law of Cosines, we have (2b+c)^2 = 20^2 + 10^2 – 2(20)(10)(cos(120)).
2b+c = 10 sqrt(7).
By the way, the answer is the magnitude of the resultant vector, not the resultant vector.

#### skipjack

Forum Staff
• pto189

#### pto189

The question itself is incorrect. A vector is either in component form or in linear combination with the given horizontal and vertical components or given magnitude (length)and direction angle. Second, you can add the magnitudes of 2 parallel vectors directly. However, to add 2 non parallel vectors, you have to use Law of Cosines.