multiplying two vectors - the dot product

• Sep 5th 2009, 08:52 PM
Mr Rayon
multiplying two vectors - the dot product
Find the dot product of the vectors $\displaystyle 3i + 3j$ and $\displaystyle 6i + 2j$ using the equation $\displaystyle u\cdot v = \left | u\left | \right |v \right |cos \theta$

Please show complete working out of how to do the above. Any attempt to do so will be appreciated.
• Sep 5th 2009, 09:16 PM
JML
To do this question in the specified way you need to find the lengths of both vectors and then the angle between the vectors. If $\displaystyle u = x i + y j$ then $\displaystyle \left| u \right| = \sqrt{x^2 + y^2}$ the same method is used for the length of v.

If the question did not specify the method you had to use for the dot product it would likely be easier to calculate the dot product as follows:

say we have $\displaystyle u = a_1 i + a_2 j$
$\displaystyle v = b_1 i + b_2 j$
then:
$\displaystyle u \cdot v = a_1 b_1 +a_2 b_2$
• Sep 5th 2009, 09:40 PM
Mr Rayon
Quote:

Originally Posted by JML
To do this question in the specified way you need to find the lengths of both vectors and then the angle between the vectors.

But how do you find the angle between the two vectors?
• Sep 5th 2009, 09:47 PM
mr fantastic
Quote:

Originally Posted by Mr Rayon
But how do you find the angle between the two vectors?

Draw the two vectors and calculate the angle that each makes with the i direction.
• Sep 5th 2009, 09:55 PM
JML
You can find the angle by noting the coordinates that the vector would end at if it started at the origin.

so let's say $\displaystyle \textbf{a} = 3i + 3j$ and $\displaystyle \textbf{b} = 6i + 2j$. So for a we get the point A =(3,3) and for b we get the point B=(6,2) . First we find the angle between the line x=0 (the i direction) and the line that passes through the origin and A, this angle = $\displaystyle \frac{\pi}{4}$ . You can do something similar for the other vector. You can then find the angle between the vectors by subtracting the angle to B from the angle to A.
• Sep 5th 2009, 10:15 PM
Mr Rayon
$\displaystyle u = 3i + 3j$and $\displaystyle v = 6i + 2j$
$\displaystyle \left | u \right | = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \left | v \right | = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}$

Okay, I've drawn a diagram...and then I find the difference between the angles:

$\displaystyle cos \theta = \frac{3}{3\sqrt{2}}$, $\displaystyle cos \theta = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}$

$\displaystyle cos\theta = \frac{6}{2\sqrt{10}}$, $\displaystyle cos\theta = \frac{6\sqrt{10}}{20} = \frac{3\sqrt{10}}{10}$

$\displaystyle \frac{3\sqrt{10}}{10} - \frac{5sqrt{2}}{10}$
$\displaystyle u\cdot v = \left |u \right | \left | v \right |cos\theta$

I am lost from here, and is the above right? Can somebody provide complete working out?

EDIT: I don't know what I'm doing here.
• Sep 5th 2009, 10:37 PM
mr fantastic
Quote:

Originally Posted by Mr Rayon
$\displaystyle u = 3i + 3j$and $\displaystyle v = 6i + 2j$
$\displaystyle \left | u \right | = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \left | v \right | = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}$

Okay, I've drawn a diagram...and then I find the difference between the angles:

$\displaystyle cos \theta = \frac{3}{3\sqrt{2}}$, $\displaystyle cos \theta = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}$

$\displaystyle cos\theta = \frac{6}{2\sqrt{10}}$, $\displaystyle cos\theta = \frac{6\sqrt{10}}{20} = \frac{3\sqrt{10}}{10}$

$\displaystyle \frac{3\sqrt{10}}{10} - \frac{5sqrt{2}}{10}$
$\displaystyle u\cdot v = \left |u \right | \left | v \right |cos\theta$

I am lost from here, and is the above right? Can somebody provide complete working out?

EDIT: I don't know what I'm doing here.

Let $\displaystyle \alpha$ be the angle between 3i + 3j and the i direction. Let $\displaystyle \beta$ be the angle between 6i + 2j and the i direction.

The angle between 3i + 3j and 6i + 2j is $\displaystyle \theta = \alpha - \beta$.

Calculate $\displaystyle \cos \theta$ using the compound angle formula. This will require that you first get $\displaystyle \sin \alpha$, $\displaystyle \cos \alpha$, $\displaystyle \sin \beta$ and $\displaystyle \cos \beta$.
• Sep 5th 2009, 10:41 PM
JML
$\displaystyle cos \theta_A = \frac{3}{3\sqrt{2}}$, $\displaystyle cos \theta_A = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}$
so $\displaystyle \theta_A = arccos \frac{\sqrt{2}}{2} = \frac{\pi}{4}$

$\displaystyle \tan \theta_b = \frac{2}{6} = \frac{1}{3}$

$\displaystyle \theta_b = arctan \frac{1}{3}$

so $\displaystyle \theta = \theta_A - \theta_B$
$\displaystyle \theta = \frac{\pi}{4} - arctan \frac{1}{3}$
• Sep 6th 2009, 03:55 AM
HallsofIvy
Quote:

Originally Posted by Mr Rayon
Find the dot product of the vectors $\displaystyle 3i + 3j$ and $\displaystyle 6i + 2j$ using the equation $\displaystyle u\cdot v = \left | u\left | \right |v \right |cos \theta$

Please show complete working out of how to do the above. Any attempt to do so will be appreciated.

Draw a picture showing $\displaystyle \vec{u}= 3\vec{i}+ 3\vec{j}$, $\displaystyle \vec{v}= 6\vec{i}+2\vec{j}$ and $\displaystyle \vec{u}- \vec{v}= -3\vec{i}+ \vec{j}$, the vector from the tip of $\displaystyle \vec{v}$ to the tip of $\displaystyle \vec{u}$.

Now you have a triangle with sides of length $\displaystyle a= |\vec{u}|= 3\sqrt{2}$, $\displaystyle b= |\vec{v}|= 2\sqrt{10}$, and $\displaystyle c= |\vec{u}-\vec{v}|= \sqrt{10}$. You can use the cosine law, $\displaystyle c^2= a^2+ b^2- 2ab \cos(C)$ to find C, the angle between the vectors.