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*}

The post Dot Product, Lengths, and Distances of Complex Vectors first appeared on Problems in Mathematics.

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 orthogonal to $\mathbf{v}_1$ and $\mathbf{v}_3$, and satisfying
\[ \mathbf{v}_2 \cdot \mathbf{v}_4 = -3 . \]

Calculate the following expressions:

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

(b) $\mathbf{v}_3 \cdot \mathbf{v}_4$.

(c) $( 2 \mathbf{v}_1 + 3 \mathbf{v}_2 – \mathbf{v}_3 ) \cdot \mathbf{v}_4 $.

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

(e) What is the distance between $\mathbf{v}_1$ and $\mathbf{v}_2$?

Solution.

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

\[ \mathbf{v}_1 \cdot \mathbf{v}_2 = \begin{bmatrix} -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} = -6 . \]

(b) $\mathbf{v}_3 \cdot \mathbf{v}_4$.

We are given that $\mathbf{v}_3$ and $\mathbf{v}_4$ are orthogonal vectors, thus
\[ \mathbf{v}_3 \cdot \mathbf{v}_4 = 0 . \]

(c) $( 2 \mathbf{v}_1 + 3 \mathbf{v}_2 – \mathbf{v}_3 ) \cdot \mathbf{v}_4 $.

First, distribute the dot product over the sum:
\[ ( 2 \mathbf{v}_1 + 3 \mathbf{v}_2 – \mathbf{v}_3 ) \cdot \mathbf{v}_4 = 2 \mathbf{v}_1 \cdot \mathbf{v}_4 + 3 \mathbf{v}_2 \cdot \mathbf{v}_4 – \mathbf{v}_3 \cdot \mathbf{v}_4 . \]

Next we use the given value for $\mathbf{v}_2 \cdot \mathbf{v}_4$, along with the given facts that $\mathbf{v}_4$ is orthogonal to both $\mathbf{v}_1$ and $\mathbf{v}_3$:
\begin{align*}
&2 \mathbf{v}_1 \cdot \mathbf{v}_4 + 3 \mathbf{v}_2 \cdot \mathbf{v}_4 – \mathbf{v}_3 \cdot \mathbf{v}_4 \\
&=2\cdot 0 +3 \cdot (-3)-0 =-9.
\end{align*}

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

The length of a general vector $\mathbf{w}$ is $\|\mathbf{w}\|:=\sqrt{ \mathbf{w}^{\trans} \mathbf{w} }$. Thus,
\[ \| \mathbf{v}_1 \| \, = \, \sqrt{ (-1)^2+0^2+2^2} \, = \, \sqrt{5} , \] \[ \| \mathbf{v}_2 \| \, = \, \sqrt{ 0^2+2^2+(-3)^2} \, = \, \sqrt{13} , \] \[ \| \mathbf{v}_3 \| \, = \, \sqrt{ 2^2+2^2+3^2 } \, = \, \sqrt{17} . \]

(e) What is the distance between $\mathbf{v}_1$ and $\mathbf{v}_2$?

The distance between the two vectors is defined to be $ \| \mathbf{v}_1 – \mathbf{v}_2 \| $. First we calculate
\[ \mathbf{v}_1 – \mathbf{v}_2 \, = \, \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} – \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} \, = \, \begin{bmatrix} -1 \\ -2 \\ 5 \end{bmatrix} . \]

Thus,
\[ \| \mathbf{v}_1 – \mathbf{v}_2 \| = \sqrt{ 1 + 4 + 25 } = \sqrt{30} . \] Click here if solved 18

The post Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors first appeared on Problems in Mathematics.

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 that this length is the distance between $\mathbf{a}$ and $\mathbf{b}$.)

Solution.

Recall that the length of a vector $\mathbf{x}$ is defined to be
\[\|\mathbf{x}\|=\sqrt{\mathbf{x}^{\trans}\mathbf{x}},\] where $\mathbf{x}^{\trans}$ is the transpose of $\mathbf{x}$.

Also, recall that the inner product of two vectors $\mathbf{x}, \mathbf{y}$ are commutative.
Namely we have
\[\mathbf{x}\cdot \mathbf{y}=\mathbf{x}^{\trans}\mathbf{y}=\mathbf{y}^{\trans}\mathbf{x}=\mathbf{y} \cdot \mathbf{x}.\]

Applying the second fact with given vectors $\mathbf{a}, \mathbf{b}$, we obtain
\[\mathbf{a}^{\trans}\mathbf{b}=\mathbf{b}^{\trans}\mathbf{a}= -\frac{1}{2}.\]

Now we compute $\|\mathbf{a}-\mathbf{b}\|^2$ as follows.
We have
\begin{align*}
\|\mathbf{a}-\mathbf{b}\|^2&=(\mathbf{a}-\mathbf{b})^{\trans}(\mathbf{a}-\mathbf{b}) \qquad \text{ (by definition of the length)}\\
&=(\mathbf{a}^{\trans}-\mathbf{b}^{\trans})(\mathbf{a}-\mathbf{b})\\
&=\mathbf{a}^{\trans}\mathbf{a}-\mathbf{a}^{\trans}\mathbf{b}-\mathbf{b}^{\trans}\mathbf{a}+\mathbf{b}^{\trans}\mathbf{b}\\
&=\|\mathbf{a}\|^2-\mathbf{a}^{\trans}\mathbf{b}-\mathbf{b}^{\trans}\mathbf{a}+\|\mathbf{b}\|^2\\
&=1-\left(-\frac{1}{2} \right)-\left(-\frac{1}{2} \right)+1\\
&=3.
\end{align*}

Since the length is nonnegative, we take the square root of the above equality and obtain
\[\|\mathbf{a}-\mathbf{b}\|=\sqrt{3}.\] Click here if solved 17

The post Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given first appeared on Problems in Mathematics.