# Dot Product, Lengths, and Distances of Complex Vectors

## Problem 689

For this problem, use the complex vectors
$\mathbf{w}_1 = \begin{bmatrix} 1 + i \\ 1 – i \\ 0 \end{bmatrix} , \, \mathbf{w}_2 = \begin{bmatrix} -i \\ 0 \\ 2 – i \end{bmatrix} , \, \mathbf{w}_3 = \begin{bmatrix} 2+i \\ 1 – 3i \\ 2i \end{bmatrix} .$

Suppose $\mathbf{w}_4$ is another complex vector which is orthogonal to both $\mathbf{w}_2$ and $\mathbf{w}_3$, and satisfies $\mathbf{w}_1 \cdot \mathbf{w}_4 = 2i$ and $\| \mathbf{w}_4 \| = 3$.

Calculate the following expressions:

(a) $\mathbf{w}_1 \cdot \mathbf{w}_2$.

(b) $\mathbf{w}_1 \cdot \mathbf{w}_3$.

(c) $((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4$.

(d) $\| \mathbf{w}_1 \| , \| \mathbf{w}_2 \|$, and $\| \mathbf{w}_3 \|$.

(e) $\| 3 \mathbf{w}_4 \|$.

(f) What is the distance between $\mathbf{w}_2$ and $\mathbf{w}_3$?

## Solution.

### (a) $\mathbf{w}_1 \cdot \mathbf{w}_2$.

$\mathbf{w}_1 \cdot \mathbf{w}_2 = \begin{bmatrix} 1+i & 1-i & 0 \end{bmatrix} \begin{bmatrix} -i \\ 0 \\ 2-i \end{bmatrix} = (1+i)(-i) + 0 + 0 = 1 – i .$

### (b) $\mathbf{w}_1 \cdot \mathbf{w}_3$.

\begin{align*} \mathbf{w}_1 \cdot \mathbf{w}_3 &= \begin{bmatrix} 1+i & 1-i & 0 \end{bmatrix} \begin{bmatrix} 2+i \\ 1-3i \\ 2i \end{bmatrix} \\ &= (1+i)(2+i) + (1-i)(1-3i) + 0 \\ &= (1 + 3i) + (-2 – 4i) \\ &= -1 – i . \end{align*}

### (c) $((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4$.

\begin{align*} ((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4 &= (2+i)( \mathbf{w}_1 \cdot \mathbf{w}_4) – (1+i) ( \mathbf{w}_2 \cdot \mathbf{w}_4 ) \\
&= (2+i) ( 2i ) – (1+i)(0) \\
&= -2 + 4i \end{align*}

Note that $\mathbf{w}_2 \cdot \mathbf{w}_4=0$ because these vectors are orthogonal.

### (d) $\| \mathbf{w}_1 \| , \| \mathbf{w}_2 \|$, and $\| \mathbf{w}_3 \|$.

For an arbitrary complex vector $\mathbf{v}$, its length is defined to be
$\| \mathbf{v} \| = \sqrt{ \overline{\mathbf{v}}^\trans \mathbf{v} } .$

Thus,
$\| \mathbf{w}_1 \| \, = \, \sqrt{ (1-i)(1+i) + (1+i)(1-i) + 0 } = \sqrt{ 2 + 2} = \sqrt{4} ,$ $\| \mathbf{w}_2 \| \, = \, \sqrt{ (i)(-i) + 0 + (2+i)(2-i) } = \sqrt{1 + 5} = \sqrt{6} ,$ $\| \mathbf{w}_3 \| \, = \, \sqrt{ (2-i)(2+i) + (1+3i)(1-3i) + (-2i)(2i) } = \sqrt{ 5 + 10 + 4} = \sqrt{19} .$

### (e) $\| 3 \mathbf{w}_4 \|$.

$\| 3 \mathbf{w}_4 \| = 3 \| \mathbf{w}_4 \| = 3\cdot 3=9$ .

### (f) What is the distance between $\mathbf{w}_2$ and $\mathbf{w}_3$?

The distance between these vectors is given by $\| \mathbf{w}_2 – \mathbf{w}_3 \|$. First we calculate this difference:
$\mathbf{w}_2 – \mathbf{w}_3 \, = \, \begin{bmatrix} -i \\ 0 \\ 2 – i \end{bmatrix} – \begin{bmatrix} 2+i \\ 1 – 3i \\ 2i \end{bmatrix} \, = \, \begin{bmatrix} -2 – 2i \\ -1 + 3i \\ 2 – 3i \end{bmatrix} .$

Now the length of the complex vector is defined to be
\begin{align*}
\| \mathbf{w}_2 – \mathbf{w}_3 \| &= \sqrt{ \left( \overline{ \mathbf{w}_2 – \mathbf{w}_3 } \right)^{\trans} \left( \mathbf{w}_2 – \mathbf{w}_3 \right) } \6pt] &= \sqrt{ \begin{bmatrix} -2 + 2i & -1 – 3i & 2 + 3i \end{bmatrix} \begin{bmatrix} -2 – 2i \\ -1 + 3i \\ 2 – 3i \end{bmatrix} } \\[6pt] &= \sqrt{ (-2+2i)(-2-2i) + (-1-3i)(-1+3i) + (2+3i)(2-3i) } \\[6pt] &= \sqrt{ 8 + 10 + 13 } \\[6pt] &= \sqrt{ 31} \end{align*} ### More from my site • Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors For this problem, use the real vectors \[ \mathbf{v}_1 = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} , \mathbf{v}_2 = \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} , \mathbf{v}_3 = \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix} . Suppose that $\mathbf{v}_4$ is another vector which is […]
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