Question

Properties of Vectors

Let $u,v$ and $w$ be vectors, and let $c$ be a scalar. Prove the given property.

$\left(c\mathbf{u}\right)·\mathbf{v}=c(\mathbf{u}·\mathbf{v})=\mathbf{u}·\left(c\mathbf{v}\right)$

Short answer

The expression $\left(c\mathbf{u}\right)·\mathbf{v}=c(\mathbf{u}·\mathbf{v})=\mathbf{u}·\left(c\mathbf{v}\right)$ is an identity

Step-by-step solution

Step 1. Given information

Expression is given as-

$\left(c\mathbf{u}\right)·\mathbf{v}=c(\mathbf{u}·\mathbf{v})=\mathbf{u}·\left(c\mathbf{v}\right)$

Step 2. Concept used

The scalar product of two vectors is commutative.

Commutative Property:

$xy=yx$

Scalar product or Dot product:

$a·b=\left|a\right|\left|b\right|\cos \theta$

Simplify LHS and RHS separately of the expression. If both are equal then expression is an identity.

Step 3. Calculation

Let $u=⟨a,b⟩\&v=⟨c,d⟩$and a constant $c$,

Simplify first part,

$\begin{array}{l}\left(cu\right)\cdot v=c⟨a,b⟩\cdot ⟨p,q⟩\\ \left(cu\right)\cdot v=⟨ca,cb⟩\cdot ⟨p,q⟩\\ \left(cu\right)\cdot v=cap+cbq\end{array}$

Simplify middle part,

$\begin{array}{l}c(u\cdot v)=c⟨a,b⟩⟨p,q⟩\\ c(u\cdot v)=c(ap+bq)\\ c(u\cdot v)=cap+cbq\end{array}$

Simplify last part,

$\begin{array}{l}u\cdot \left(cv\right)=⟨a,b⟩(c⟨p,q⟩)\\ u\cdot \left(cv\right)=⟨a,b⟩⟨cp,cq⟩\\ u\cdot \left(cv\right)=acp+bcq\end{array}$

All three terms are equal. Hence, expression is an identity.