# Inner Product, Norm, and Orthogonal Vectors

## Problem 162

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 $\mathbf{u_1}=\mathbf{u_2}+a\mathbf{u}_3$.

(The Ohio State University, Linear Algebra Exam Problem)

Contents

## Hint.

Recall the following definitions.

• The inner product (dot product) of two vectors $\mathbf{v}_1, \mathbf{v}_2$ is defined to be
$\mathbf{v}_1\cdot \mathbf{v}_2 :=\mathbf{v}^{\trans}_1\mathbf{v}_2.$
• Two vectors $\mathbf{v}_1, \mathbf{v}_2$ are orthogonal if the inner product
$\mathbf{v}_1\cdot \mathbf{v}_2=0.$
• The norm (length, magnitude) of a vector $\mathbf{v}$ is defined to be
$||\mathbf{v}||=\sqrt{\mathbf{v}\cdot \mathbf{v}}.$

## Solution.

We first express the given conditions in term of inner products (dot products).
Since $\mathbf{u}_1$ and $\mathbf{u}_2$ are orthogonal, the inner product
$\mathbf{u}_2\cdot \mathbf{u}_1=\mathbf{u}_1 \cdot \mathbf{u}_2=0. \tag{a}$

Also, since the norm of $\mathbf{u}_2$ is $4$, we obtain
$\mathbf{u}_2\cdot\mathbf{u}_2=||\mathbf{u}_2||^2=16. \tag{b}$ The last condition can be written as
$\mathbf{u}_2\cdot \mathbf{u}_3=\mathbf{u}_2^{\trans}\mathbf{u}_3=7 \tag{c}.$

Now we compute the inner product $\mathbf{u}_2$ and $\mathbf{u}_1$.
\begin{align*}
0\stackrel{(a)}{=}&\mathbf{u}_2\cdot \mathbf{u}_1 =\mathbf{u}_2\cdot (\mathbf{u}_2+a\mathbf{u}_3)\\
&=\mathbf{u}_2\cdot \mathbf{u}_2+a\mathbf{u}_2\cdot \mathbf{u}_3\\
&\stackrel{(b), (c)}{=}16+7a.
\end{align*}
Therefore, solving this we obtain
$a=-\frac{16}{7}.$

### More from my site

#### You may also like...

##### Give a Formula for a Linear Transformation if the Values on Basis Vectors are Known

Let $T: \R^2 \to \R^2$ be a linear transformation. Let \[ \mathbf{u}=\begin{bmatrix} 1 \\ 2 \end{bmatrix}, \mathbf{v}=\begin{bmatrix} 3 \\ 5...

Close