# Thread: Vectors (Dot product?)

1. ## Vectors (Dot product?)

Given that the two vectors u=(2a+b) and v=(4a-3b) are perpendicular and that |a|=3 and |b|=6, then find the angle between a and b.

So I pretty sure i need to set the dot product of u and v = to zero. Then there is probably some sort of way to rearange it, of which I am not sure.

2. Originally Posted by Mtl
Given that the two vectors u=(2a+b) and v=(4a-3b) are perpendicular and that |a|=3 and |b|=6, then find the angle between a and b.

So I pretty sure i need to set the dot product of u and v = to zero. Then there is probably some sort of way to rearange it, of which I am not sure.
Hello,

the definition of the dot product says:

$\vec a \cdot \vec b = |\vec a| \cdot |\vec b| \cdot \cos(\theta)$ and therefore:

$\cos(\theta)=\frac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}$

From your problem you know:

$(2\vec a + \vec b) \cdot (4 \vec a - 3\vec b) = 0$..... Expand the brackets:

$8(\vec a)^2 - 6\vec a \vec b + 4\vec a \vec b -3(\vec b)^2 = 0$

Since $(\vec a)^2 = 9$ and $(\vec b)^2 = 36$

the equation becomes:

$-2(\vec a \cdot \vec b) = 36$ and therefore $\vec a \cdot \vec b = -18$

Now plug in all values into the formula to calculate the angle:

$\cos(\theta) = \frac{-18}{3 \cdot 6}=-1~\implies~\theta = 180^\circ$

3. Thanks, well done in explaining the solution.

4. Originally Posted by Mtl
Thanks, well done in explaining the solution.
Hi,

thanks for your kind reply BUT: your problem is a little bit more tricky than it seems to be:

1. According to my result the vectors $\vec a$ and $\vec b$ are pointing in opposite directions, that means they are collinear. And therefore any linear combinations of $\vec a$ and $\vec b$ must be parallel. Seems to be a contradiction to the given conditions.

2. Considering the directions of $\vec a$ and $\vec b$ and the length of $\vec a$ and $\vec b$ then

$\vec u = 0$

and everything is nice and peaceful again.