Find perimeter and angle of triangle using three 3d vectors . Given the following, three vectors:
$$\vec{a} = 3\mathrm{i} - 2\mathrm{j} + 5\mathrm{k}\\\vec{b} = \mathrm{i} - 6\mathrm{j} + 6\mathrm{k}\\\vec{c} = 2\mathrm{i} + 3\mathrm{j} - \mathrm{k},\\$$
find the perimeter and angles of the triangle using three 3D vectors.
Studying for exam and this question is giving me difficulty. My attempt at finding the perimeter is finding length of vectors and then adding them together. E.g $|A|^2=9+4+25=38$ and so on, correct ? How would I go about finding the perimeter and angles of this triangle.
 A: The side vectors of the triangle are given by the differences of the position vectors of the vertices. For example
$$\vec a -\vec b = 2i+4j-k$$
is one of the sides whose length is $\sqrt{4+16+1}=\sqrt{21}.$
Can you do the same with the other two edges?
The following figure may help to better understand the solution:

Regarding the angles: Let's calculate another side vector:
$$\vec a -\vec c=i-5j+6k.$$
The length of this vector is $\sqrt{62}.$
So,
$$\vec x=\frac{\vec a- \vec b}{\sqrt{21}}$$
and
$$\vec y=\frac{\vec a- \vec c}{\sqrt{62}}$$
are two unit vectors. The scalar product of these two vectors will give you the cosine of the angle of the two corresponding edges, the angle at $\vec a$.
Can you do that with the other two pairs of edges?
A: The vectors given are positions of the triangle's vertices. Vector $$\vec a = 3i−2j+5k$$ represents a vertex $A$ whose coordinates are $(3, -2, 5)$.
Use $$distance = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$$ for each pair of vertices to calculate the triangle sides' lengths, then add them to obtain a perimeter.
Once you know the sides, use the law of cosines to calculate angles.
A: Perimeter is the sum of all sides.. Add all three sides and find the magnitude
i.e |A|+|B|+|C|=SOme number then again find the magnitude of this number you will find the answer and for angle use law of cosine
