# Dot product and unit vectors

• Apr 30th 2010, 08:46 AM
Kakariki
Dot product and unit vectors
Hey,

I am working through work on the dot product and came across this question:

Let a be a unit vector and let |b| = 4. Let the angle between these two vectors be 30 degrees. Calculate (2a - b) DOT (5a + 3b).

*Don't know how to make the arrow above the letters, pretend it is above all a's and b's. Also, the DOT is how I am going to say dot product like the formula for the dot product of two vectors with an angle theta between them is: u DOT v = |u| |v| cos(theta).

I am unsure of how to proceed with this question. I am assuming that I put this through the DOT formula. But am unsure of how to set this up.

Help?
• Apr 30th 2010, 09:01 AM
Plato
$$\vec{a}\cdot\vec{b}$$ gives $\displaystyle \vec{a}\cdot\vec{b}$.

$\displaystyle (2\vec{a}-\vec{b})\cdot (5\vec{a}+3\vec{b})=10\vec{a}\cdot\vec{a}+\vec{a}\ cdot\vec{b}-3\vec{b}\cdot\vec{b}$

$\displaystyle \cos(\theta)=\frac{\vec{a}\cdot\vec{b}}{||a||~||b| |}$.
• Apr 30th 2010, 09:07 AM
shenanigans87
Let a be a unit vector and let |b| = 4. Let the angle between these two vectors be 30 degrees. Calculate (2a - b) DOT (5a + 3b).

If a is a unit vector, let it equal {1,0,0}

then by taking the sin and cos of 30, we know that b would be $\displaystyle (\frac{\sqrt{3}}{2},\frac{1}{2},0)$

but since |b|=4, b is really equal to

$\displaystyle 4*(\frac{\sqrt{3}}{2},\frac{1}{2},0)=(2\sqrt{3},2, 0)$

now that you know a and b, you should be able to do those dot products pretty easily.
• Apr 30th 2010, 09:55 AM
Kakariki
Quote:

Originally Posted by shenanigans87
Let a be a unit vector and let |b| = 4. Let the angle between these two vectors be 30 degrees. Calculate (2a - b) DOT (5a + 3b).

If a is a unit vector, let it equal {1,0,0}

then by taking the sin and cos of 30, we know that b would be $\displaystyle (\frac{\sqrt{3}}{2},\frac{1}{2},0)$

but since |b|=4, b is really equal to

$\displaystyle 4*(\frac{\sqrt{3}}{2},\frac{1}{2},0)=(2\sqrt{3},2, 0)$

now that you know a and b, you should be able to do those dot products pretty easily.

Okay, using those numbers I get an answer of $\displaystyle 2\sqrt{3} - 38$

How did you know to use the coordinates you used for a and b?
• Apr 30th 2010, 10:12 AM
shenanigans87
All you were told is that a is a unit vector and b is 30 degrees away with a magnitude of 4. You can pick any two vectors that satisfy that, I just picked the easiest.

And the answer is actually $\displaystyle 2\sqrt{3}-38$
• Apr 30th 2010, 10:14 AM
Kakariki
Quote:

Originally Posted by shenanigans87
All you were told is that a is a unit vector and b is 30 degrees away with a magnitude of 3. You can pick any two vectors that satisfy that, I just picked the easiest.

Oh! Could you list 2 or 3 others that satisfy the question? Just to be sure that I can understand completely how to choose the vectors for a question like this.

Thank you so much!