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.

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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 $d_1 \cdot d_3$.


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