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)

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
\[a=-\frac{16}{7}.\]

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