Modulus of vectors squared

Question

Recently, I saw someone write: $|OC - OB|^2 = |OC^2 - 2OC\cdot OB + OB^2|$ where $OC$ and $OB$ are both vectors.

I was wondering if this is the same thing or not?

Thank you in advance!

Answer 1

From the definition of modulus and the properties of the dot product:

$$ \begin{align} |OC-OB|^2 & = (OC-OB)\cdot (OC-OB) \\ & = OC \cdot OC - OC \cdot OB - OB \cdot OC + OB \cdot OB \\ & = |OC|^2 - 2\; OC \cdot OB + |OB|^2 \end{align} $$

Answer 2

Hint:

See the figure:

$ \vec {AB}-\vec{AC}=\vec {BC} $ is the side of the triangle $ABC$ opposite to the angle $A$, so the identity $$ |\vec AB-\vec AC|^2=|\vec AB|^2 -2\vec{AB}\cdot \vec{AC}+|\vec AC|^2 $$

is the law of cosines applied to the triangle $ABC$, with the geometric interpretation of the dot product. (Note that the use of the modulus is wrong in OP).

Answer 3

Let's go from the basics. Take $OC=\sum_{i=1}^na_ix_i, \ \ OB=\sum_{i=1}^nb_ix_i$ where $\{x_i\}$ form the orthogonal basis of the space. $$OC-OB=\sum_{i=1}^n(a_i-b_i)x_i\\\implies |OC-OB|^2=(\sum_{i=1}^n(a_i-b_i))^2$$ Now $$OC^2-2OC\cdot OB+OB^2\\=\sum_{i=1}^na_i^2-2\sum_{i=1}^na_ib_i+\sum_{i=1}^nb_i^2\\=(\sum_{i=1}^n(a_i-b_i))^2$$