1. ## Vectors question

I'm not sure where you'd ask vector questions, but since this one involved $\displaystyle \cos$ I decided to ask it here:

Two vectors, a and b, have a common starting point with an angle of 120 degrees between them. The vectors are such that |a| = 3 and |b| = 4.

Calculate |a+b|

The way I thought of it was |a+b| = \sqrt{3^2+4^2-2(3)(4)(\cos{60})} = \sqrt {13}

However, the answer in the book is \sqrt {3}

2. Originally Posted by IanCarney
I'm not sure where you'd ask vector questions, but since this one involved $\displaystyle \cos$ I decided to ask it here:

Two vectors, a and b, have a common starting point with an angle of 120 degrees between them. The vectors are such that |a| = 3 and |b| = 4.

Calculate |a+b|

The way I thought of it was |a+b| = \sqrt{3^2+4^2-2(3)(4)(\cos{60})} = \sqrt {13}

However, the answer in the book is \sqrt {3}
you need to use the dot (inner) product

$\displaystyle |\mathbf{a}+\mathbf{b}|^2=(\mathbf{a}+\mathbf{b}) \cdot (\mathbf{a}+\mathbf{b})$

Now expand using the definition of the dot product and remember the geometric definition is

$\displaystyle \mathbf{a}\cdot \mathbf{b}= |\mathbf{a}||\mathbf{b}|\cos(\theta)$

3. Originally Posted by TheEmptySet
you need to use the dot (inner) product

$\displaystyle |\mathbf{a}+\mathbf{b}|^2=(\mathbf{a}+\mathbf{b}) \cdot (\mathbf{a}+\mathbf{b})$

Now expand using the definition of the dot product and remember the geometric definition is

$\displaystyle \mathbf{a}\cdot \mathbf{b}= |\mathbf{a}||\mathbf{b}|\cos(\theta)$
Sorry, I haven't learned the dot product rule (yet). The way I've been taught so far has just been to "connect" the vectors and use the cosine law.

4. Originally Posted by IanCarney
Two vectors, a and b, have a common starting point with an angle of 120 degrees between them. The vectors are such that |a| = 3 and |b| = 4.
Calculate |a+b|
This is a result of the parallelogram law.
If $\displaystyle \theta$ is the angle between $\displaystyle \mathbf{\vec{a}}~\&~\mathbf{\vec{b}}$ then
$\displaystyle |\mathbf{\vec{a}}+\mathbf{\vec{b}}|^2=|\mathbf{a}| ^2+|\mathbf{\vec{b}}|^2-2|\mathbf{\vec{a}}||\mathbf{\vec{b}}|\cos(\pi -\theta)$

5. Originally Posted by Plato
This is a result of the parallelogram law.
If $\displaystyle \theta$ is the angle between $\displaystyle \mathbf{\vec{a}}~\&~\mathbf{\vec{b}}$ then
$\displaystyle |\mathbf{\vec{a}}+\mathbf{\vec{b}}|^2=|\mathbf{a}| ^2+|\mathbf{\vec{b}}|^2-2|\mathbf{\vec{a}}||\mathbf{\vec{b}}|\cos(\pi -\theta)$
Would you agree with my answer then? |a+b| =

6. Originally Posted by IanCarney
Would you agree with my answer then? |a+b| =
Draw a sketch to scale for clarification (see attachment)

... and $\displaystyle \sqrt{13} \approx 3.60555$

It seems to me that there is a typo in your book.

7. Another way to do this is to note that you have two sides of a triangle, having lengths 3 and 4 with angle between them 180- 120= 60 degrees. The length of the third side, which is |a+ b| can be calculated using the cosine law.
$\displaystyle \sqrt{13}$
is correct.

8. Originally Posted by earboth
Draw a sketch to scale for clarification (see attachment)

... and $\displaystyle \sqrt{13} \approx 3.60555$

It seems to me that there is a typo in your book.
Thanks! I had done that but the answer in the book threw me off.

Another question: Calculate the angle between vector a and vector a+b.

I would do sin60/sqrt{13}=sinx/4, where x is the angle, right? The answer in the book appears to be another typo, so I just wanted some clarification.

9. Originally Posted by IanCarney
Thanks! I had done that but the answer in the book threw me off.

Another question: Calculate the angle between vector a and vector a+b.

I would do sin60/sqrt{13}=sinx/4, where x is the angle, right? The answer in the book appears to be another typo, so I just wanted some clarification.
You have three sides and one angle of a triangle. Use Law of Sines. What answer did you get?

-Dan

10. Originally Posted by IanCarney
Thanks! I had done that but the answer in the book threw me off.

Another question: Calculate the angle between vector a and vector a+b.

I would do sin60/sqrt{13}=sinx/4, where x is the angle, right? The answer in the book appears to be another typo, so I just wanted some clarification.
$\displaystyle \vec r = \vec a + \vec b$
$\displaystyle \angle(\vec a, \vec r) = \beta$

2. Y0u know:

$\displaystyle |\vec a| = 3$ , $\displaystyle |\vec b| = 4$ , $\displaystyle |\vec r| = \sqrt{13}$

3. According to the Cosine rule you have:

$\displaystyle (\vec b)^2= (\vec a)^2+(\vec r)^2 - 2 \cdot \vec a \cdot \vec r \cdot \cos(\beta)$

Determine $\displaystyle \cos(\beta)$ and consequently ß.

11. Originally Posted by earboth
$\displaystyle \vec r = \vec a + \vec b$
$\displaystyle \angle(\vec a, \vec r) = \beta$

2. Y0u know:

$\displaystyle |\vec a| = 3$ , $\displaystyle |\vec b| = 4$ , $\displaystyle |\vec r| = \sqrt{13}$

3. According to the Cosine rule you have:

$\displaystyle (\vec b)^2= (\vec a)^2+(\vec r)^2 - 2 \cdot \vec a \cdot \vec r \cdot \cos(\beta)$

Determine $\displaystyle \cos(\beta)$ and consequently ß.
Yeah, you can do it that way too. (I just like the Law of Sines better in this circumstance. But no worries.)

-Dan