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If the length of vector ${\bf a}$ is $a=|{\bf a}|$, then we have $a^2={\bf a}\cdot {\bf a}$.
We have $\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}={\bf b} - {\bf a}$.
It might be easier to consider $(AB^2+CD^2) - (AC^2+BD^2)$ and show that this is equal to $0$ rather than trying to show that $AB^2+CD^2 = AC^2+BD^2$.
A tetrahedron has two identical equilateral triangles faces, of side length 1 unit. The other two faces are right angled isosceles triangles. Find the exact volume of the tetrahedron.
Can you prove that in every tetrahedron there is a vertex where the three edges meeting at that vertex have lengths which could be the sides of a triangle?
ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square.