how to compare slope of two lines and find the angles in 3d

Question

I have three points $A(x_1,y_1,z_1)$ ,$ B(x_2,y_2,z_2)$ and $C(x_3,y_3,z_3)$ and I want to compare the slopes of AB and BC and check the angles between them.
What should I do ?
Thanks in advance .

Answer 1

Hint 1: Find the direction cosines of $AB$ and $BC$. Then take their dot product. This will give the cosine of the angle between $AB$ and $BC$. $$$$ Hint 2: The Direction Cosines of $PQ$, $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, are $\left(\pm\frac{x_2-x_1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}},\pm\frac{y_2-y_1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}},\pm\frac{z_2-z_1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}\right)$ $$$$ Edit: The Dot Product works iff the lines intersect.

Answer 2

Lines in dimension three do not have to intersect. We find their direction vectors to see if they are parallel.

Note that $$ \vec {AB} = < x_2-x_1, y_2 - y_1, z_2-z_1>$$ and $$ \vec {BC} = < x_3-x_2, y_3 - y_2, z_3-z_2>$$

These vectors are the direction vectors are your lines.

The dot product $$ \vec {AB}.\vec {BC}$$ along with the norms of these vectors determine the angle between the lines if they intersect at all. $$ \cos\theta = \frac { \vec {AB}.\vec {BC}}{||AB||.||BC||}$$