Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors

Linear algebra problems and solutions

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 $\mathbf{v}_2$?

 
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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,
\[ \| \mathbf{v}_1 – \mathbf{v}_2 \| = \sqrt{ 1 + 4 + 25 } = \sqrt{30} . \]


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