# 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*} Sponsored Links ### 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 […]
• Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are $\|\mathbf{a}\|=\|\mathbf{b}\|=1$ and the inner product $\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.$ Then determine the length $\|\mathbf{a}-\mathbf{b}\|$. (Note […]
• Eigenvalues and Eigenvectors of The Cross Product Linear Transformation We fix a nonzero vector $\mathbf{a}$ in $\R^3$ and define a map $T:\R^3\to \R^3$ by $T(\mathbf{v})=\mathbf{a}\times \mathbf{v}$ for all $\mathbf{v}\in \R^3$. Here the right-hand side is the cross product of $\mathbf{a}$ and $\mathbf{v}$. (a) Prove that $T:\R^3\to \R^3$ is […]
• Unit Vectors and Idempotent Matrices A square matrix $A$ is called idempotent if $A^2=A$. (a) Let $\mathbf{u}$ be a vector in $\R^n$ with length $1$. Define the matrix $P$ to be $P=\mathbf{u}\mathbf{u}^{\trans}$. Prove that $P$ is an idempotent matrix. (b) Suppose that $\mathbf{u}$ and $\mathbf{v}$ be […]
• Find the Inverse Matrix of a Matrix With Fractions Find the inverse matrix of the matrix $A=\begin{bmatrix} \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\[6 pt] \frac{6}{7} &\frac{2}{7} &-\frac{3}{7} \\[6pt] -\frac{3}{7} & \frac{6}{7} & -\frac{2}{7} \end{bmatrix}.$   Hint. You may use the augmented matrix […]
• Inner Product, Norm, and Orthogonal Vectors Let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ are vectors in $\R^n$. Suppose that vectors $\mathbf{u}_1$, $\mathbf{u}_2$ are orthogonal and the norm of $\mathbf{u}_2$ is $4$ and $\mathbf{u}_2^{\trans}\mathbf{u}_3=7$. Find the value of the real number $a$ in […]
• Equivalent Conditions to be a Unitary Matrix A complex matrix is called unitary if $\overline{A}^{\trans} A=I$. The inner product $(\mathbf{x}, \mathbf{y})$ of complex vector $\mathbf{x}$, $\mathbf{y}$ is defined by $(\mathbf{x}, \mathbf{y}):=\overline{\mathbf{x}}^{\trans} \mathbf{y}$. The length of a complex vector […]
• Orthonormal Basis of Null Space and Row Space Let $A=\begin{bmatrix} 1 & 0 & 1 \\ 0 &1 &0 \end{bmatrix}$. (a) Find an orthonormal basis of the null space of $A$. (b) Find the rank of $A$. (c) Find an orthonormal basis of the row space of $A$. (The Ohio State University, Linear Algebra Exam […]

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### How to Obtain Information of a Vector if Information of Other Vectors are Given

Let $A$ be a $3\times 3$ matrix and let \[\mathbf{v}=\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}...

Close