# Question 05482

Mar 29, 2015

Yes, it can.

Note- For denoting vector quantities, I will bold the letters. The unbolded letters will be the magnitude, or scalar quantities, of the vectors.

Let's first define the scalar (or dot) product. The scalar product of two vectors can either defined as:

(1) A $\cdot$ B = AB$\cos \left(\theta\right)$ with theta being the angle between the two vectors and A and B being the magnitudes of vectors A and B.

or

(2) For A= < a1,a2,a3 > and B=< b1,b2,v3 >, A $\cdot$ B = (a1b1)+(a2b2)+(a3*b3)

For the case of (1), the scalar product will never be negative due to the magnitudes of the vectors since the magnitude is always positive. However, the scalar product will be negative when $\cos \left(\theta\right)$ is negative. This will happen when $\frac{\pi}{2} < \theta < \frac{3 \pi}{2}$ or, in degrees, 90° < theta < 270° #. For example, if the angle between two vectors of magnitude 5 and 2 respectively is $\frac{2 \pi}{3}$, the scalar product will be $5 \cdot 2 \cdot \cos \left(\frac{2 \pi}{3}\right) = 10 \cdot - 0.5 = - 5$

An example of a negative scalar product for the case of (2) would be if you had vector A= <2,4,-5> and B=<-1,2,3>.

A $\cdot$ B$= \left(- 2\right) + \left(8\right) + \left(- 15\right) = - 9$