# Finding dot product between two vectors with constraints

I wanted to compute the scalar product /dot product b/w two vectors with some conditions. There are three unit vectors in three-dimensional space. Let the unit vectors be, $$\vec{d}_{1},\vec{d}_{2},\vec{d}_{3}$$

If I know the dot product between two adjacent vectors, $$\vec{d}_{1} \cdot \vec{d}_{2}=\vec{d}_{2} \cdot \vec{d}_{3} = C ( say, C=\cos(60))$$ Is it possible to compute the dot product between $$\vec{d}_{1}$$ and $$\vec{d}_{3}$$, $$\vec{d}_{1} \cdot \vec{d}_{3}$$ One last question if I have a series of vectors, $$\vec{d}_{1},\vec{d}_{2} \cdots \vec{d}_{n}$$ if we know the dot product between nearest neighbor vectors, $$\vec{d}_{i} \cdot \vec{d}_{i+1} = C$$, constant, Can we compute the dot product $$\vec{d}_{i} \cdot \vec{d}_{j}$$ ?

No you can't. Basically you are asking if the angle between $$d_1$$ and $$d_3$$ is a constant if the angle between $$d_1$$ with $$d_2$$ and $$d_2$$ with $$d_3$$ is a constant.
In this image you can see that the angle between $$d_1$$ and $$d_2$$ is a constant. Furthermore the red vectors are possible solutions for $$d_3$$. You can see that there are multiple solutions and you can also see that the angle between $$d_1$$ and $$d_3$$ is different for each of them.
You could also think about it this way: Saying that a $$v_2 \cdot v_3 = konst.$$ has an infinite amount of solutions. In fact these solutions all lay on a cone around $$d_2$$. You can take any vector $$d_3$$ that satisfies your scalar product, rotate it around $$d_2$$ and you still end up with a vector that satisfies that condition. But because rotating any vector around $$d_2$$ changes the angle to the $$d_1$$ vector (unless $$d_1 = \alpha d_2$$), you cannout compute $$d_1 \cdot d_3$$.