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.
Best Answer
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:
$$\vec a -\vec c=i-5j+6k.$$
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?