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Vectors, EXT2 EQ-Bank 21

Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in the plane, where \(\mathbf{v}\,\neq\, \mathbf{0}\).

For every real number \(t\), let  \(P(t)=\abs{\mathbf{u} -t \mathbf{v}}^2\).

  1. Show that  \(P(t)=\abs{\mathbf{u}}^2-2 t( \mathbf{u} \cdot \mathbf{v} )+t^2 \abs{\mathbf{v}}^2\).   (1 mark)

  2. Show that \(P(t)\) has a minimum value at  \(t=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\).   (2 marks)

  3. Hence prove the Cauchy-Schwarz inequality,  \(\abs{\mathbf{u}\cdot\mathbf{v}} \leq \abs{\mathbf{u}}\abs{\mathbf{v}}\).   (3 marks)

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a.    \(\text{See Worked Solutions}\)

b.    \(\text{See Worked Solutions}\)

c.    \(\text{See Worked Solutions}\)

Show Worked Solution
a.     \(P(t)\) \(=\abs{\mathbf{u} -t \mathbf{v}}^2\)
    \(=( \mathbf{u} -t \mathbf{v} ) \cdot( \mathbf{u} -t\mathbf{v} )\)
    \(= \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot(-t \mathbf{v} )+(-t \mathbf{v}) \cdot \mathbf{u} +t^2(\mathbf{v} \cdot \mathbf{v} )\)
    \(=\abs{\mathbf{u}}^2-2 t( \mathbf{u} \cdot \mathbf{v} )+t^2\abs{\mathbf{v}}^2\)

 

b.    \(\text{Find}\ t\ \text{when}\ \ \dfrac{d P}{d t}=0:\)

\(2 t\abs{\mathbf{v}}^2-2 \mathbf{u} \cdot \mathbf{v}\) \(=0\)  
\(2 t\abs{\mathbf{v}}^2\) \(=2 \mathbf{u} \cdot \mathbf{v}\)  
\(t\) \(=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\)  

 

\(\dfrac{d^2 P}{d t^2}=2\abs{\mathbf{v}}^2>0\)

\(\therefore\ \text{Min SP at}\ \ t=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\)
 

c.    \(P(t)=\abs{\mathbf{u} -t \mathbf{v}}^2\ \ \Rightarrow\ \ P(t) \geq 0\)

\(P\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right)\) \(=\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right)^2\abs{\mathbf{v}}^2-2\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right) \mathbf{u} \cdot \mathbf{v} +\abs{\mathbf{u}}^2\)
  \(=-\dfrac{( \mathbf{u} \cdot \mathbf{v} )^2}{\abs{\mathbf{v}}^2}+\abs{\mathbf{u}}^2\)

 

\(0\) \( \leq -\dfrac{( \mathbf{u} \cdot \mathbf{v} )^2}{\abs{\mathbf{v}}^2}+\abs{\mathbf{u}}^2\)  
\((\mathbf{u} \cdot \mathbf{v} )^2\) \(\leq  \abs{\mathbf{u}}^2\abs{\mathbf{v}}^2\)  
\(\abs{ \mathbf{u} \cdot \mathbf{v} }\) \(\leq\abs{\mathbf{u}}\abs{\mathbf{v}}\ \ \text{(i.e. the Cauchy-Schwarz inequality)}\)