Given the following, three vectors:
\begin{align*} \vec{a}& = 3i−2j+5k, \\ \vec{b}& =i−6j+6k, \\ \vec{c}& =2i+3j−k, \\ \end{align*}
find the angles between sides of triangle and coordinate planes.
I calculated the sides to be $4.58 , 11.45$ and $7.87$. I also calculated all angles to be $17.4^{\circ}, 131.68^{\circ}$ and $30.91^{\circ}$ .
The last part of this questions requires me find the projections of the triangle on the coordinates plane
And also
Find the volume of the skewed parallelepiped with edges given by the lines $OA, OB, OC$ ($O$ is origin)
HINT 1
Given two points $A=(3,-2,5)$ and $B=(1,-6,6)$ give the unit vector parallel to the straight line $AB$. Considering the given points a possible direction vector of the straight line in question is $$\vec a-\vec b=2\vec i+4\vec j -1\vec k$$ where $\vec a$ and $\vec b$ are the position vectors belonging to the two $A$ and $B$. The best is to use a unit vector for position vector. So, let
$$\vec d=\frac{2\vec i+4\vec j -1\vec k}{\sqrt{21}}$$
be our choice.
The next problem is to find the angle of this straight line and, say, the $xy$ plane. In order to find this angle we have to project the line onto tha $xy$ plane. This is easy. We only have to omit the $\vec k$ component. The corresponding unit direction vector is
$$\vec d_{xy}=\frac{2\vec i+4\vec j }{\sqrt{20}}.$$
Now, the cosine of the angle of the two straights is the the scalar product of the normalized direction vectors:
$$\cos(\alpha)=\left(\frac{2\vec i+4\vec j -1\vec k}{\sqrt{21}}\right)\cdot \left(\frac{2\vec i+4\vec j }{\sqrt{20}}\right)=\frac{20}{\sqrt{420}}.$$ Then
$$\alpha\approx 12.6^{\circ}.$$
This angle is the angle between the $AB$ side of the triangle and the $xy$ plane.
The Volume of the Parallelepiped
Regarding the parallelepiped consider the following figure.
In order to calculate the volume of our parallelepiped we need only the following vectors]2: $$\vec c =2\vec i+3\vec j-\vec k,$$ $$\color{red}{\vec b-\vec a}=-2\vec i-4\vec j+\vec k,$$ and $$\color{blue}{\vec c-\vec a}=-\vec i+5\vec j-6\vec k$$
Given these vectors, the volume is the absolute value of the following "scalar triple product"
$$V=|\vec a\cdot (\color{red}{\vec b-\vec a})\times (\color{blue}{\vec c-\vec a}))|.$$
Now, $$(\color{red}{\vec b-\vec a})\times (\color{blue}{\vec c-\vec a})= \begin{vmatrix} \ \vec i& \ \vec j & \vec k\\ -2&-4&\ 1\\ -1&\ 5&-6 \end{vmatrix}=19\vec i -13\vec j -14\vec k$$
and $$V=|\vec a\cdot (19\vec i -13\vec j -14\vec k)|=|(3\vec i-2 \vec j+5\vec k)\cdot(19\vec i -13\vec j -14\vec k)|=13 \ \text {unit}^3.$$
($\color{red}{\text{My numerical calculations have to be triple checked!!!}}$)