Question

Given that the points A=(-1,4,0), B=(2,1,1), C=(2,-2,1), D=(-1,1,0) are four vertices of the parallelogram ABCD, determine the angles by using the dot product

Answer 1

GIVEN

The coordinates of the vertices of the parallelogram are given to be:

$A=\left(-1,4,0\right),B=\left(2,1,1\right),C=\left(2,-2,1\right),D=\left(-1,1,0\right)$

SOLUTION

Recall the formula used to calculate the angle between two vectors:

$\cos\theta=\frac{A\cdot B}{|A||B|}$

where A.B is the dot product of the vectors A and B.

The angle between vectors A and B can be calculated as follows:

$\begin{gathered} A=(-1,4,0) \\ B=(2,1,1) \\ \therefore \\ A\cdot B=(-1\times2)+(4\times1)+(0\times1)=2 \\ |A|=\sqrt{(-1)^2+4^2+0^2}=\sqrt{1+16}=\sqrt{17} \\ |B|=\sqrt{2^2+1^2+1^2}=\sqrt{4+1+1}=\sqrt{6} \end{gathered}$

Therefore:

$\begin{gathered} \cos\theta=\frac{2}{\sqrt{17}\sqrt{6}} \\ \therefore \\ \theta=78.6\degree \end{gathered}$

The same pattern is followed for the other angles.

Hence, the angles are given to be:

$\begin{gathered} Between\text{ }BC\Rightarrow65.9\degree \\ Between\text{ }CD\Rightarrow160.5\degree \\ Between\text{ }AD\Rightarrow31.0\degree \end{gathered}$