Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors
Problem 687
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 $\mathbf{v}_2$?
Sponsored Links
Contents
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,
\[ \| \mathbf{v}_1 – \mathbf{v}_2 \| = \sqrt{ 1 + 4 + 25 } = \sqrt{30} . \]
Add to solve later
Sponsored Links
More from my site
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 […]
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 […]
Sum of Squares of Hermitian Matrices is Zero, then Hermitian Matrices Are All Zero
Let $A_1, A_2, \dots, A_m$ be $n\times n$ Hermitian matrices. Show that if
\[A_1^2+A_2^2+\cdots+A_m^2=\calO,\]
where $\calO$ is the $n \times n$ zero matrix, then we have $A_i=\calO$ for each $i=1,2, \dots, m$.
Hint.
Recall that a complex matrix $A$ is Hermitian if […]
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 […]- Dot Product, Lengths, and Distances of Complex Vectors
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 […]
- Eigenvalues of a Hermitian Matrix are Real Numbers
Show that eigenvalues of a Hermitian matrix $A$ are real numbers.
(The Ohio State University Linear Algebra Exam Problem)
We give two proofs. These two proofs are essentially the same.
The second proof is a bit simpler and concise compared to the first one.
[…]
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 […]- Prove the Cauchy-Schwarz Inequality
Let $\mathbf{a}, \mathbf{b}$ be vectors in $\R^n$.
Prove the Cauchy-Schwarz inequality:
\[|\mathbf{a}\cdot \mathbf{b}|\leq \|\mathbf{a}\|\,\|\mathbf{b}\|.\]
We give two proofs.
Proof 1
Let $x$ be a variable and consider the length of the vector […]
