## Precalculus (6th Edition) Blitzer

There can be many vectors orthogonal to a given vector $\mathbf{v}$. Two possible answers are $\mathbf{u}=5\mathbf{i}+2\mathbf{j}$ and $\mathbf{u}=10\mathbf{i}+4\mathbf{j}$.
Let the vector orthogonal to $\mathbf{v}$ be $\mathbf{u}$. And it is expressed as $\mathbf{u}=a\mathbf{i}+b\mathbf{j}$ …… (1) For $\mathbf{u}$ and $\mathbf{v}$ to be orthogonal, their dot product must be zero, therefore $\mathbf{u}\cdot \mathbf{v}=-2a+5b$ …… (2) Keeping the dot product obtained in equation (2) as zero, we get $5b=2a$ …… (3) Equation (2) is a relation between $b$ and $a$ ; for different sets of $b$ and $a$, there will be different vectors orthogonal to $\mathbf{v}$. Let us put $a=5$ in equation (3) to get \begin{align} & a=5 \\ & b=2 \end{align} So, substitute these values of $b$ and $a$ in equation (1) to get $\mathbf{u}=5\mathbf{i}+2\mathbf{j}$ Hence, The vector $\mathbf{u}=5\mathbf{i}+2\mathbf{j}$ is orthogonal to $\mathbf{v}=-2\mathbf{i}+5\mathbf{j}$. Let us put $a=10$ in equation (3) to get \begin{align} & a=10 \\ & b=4 \end{align} So, substitute these values of $b$ and $a$ in equation (1) to get $\mathbf{u}=10\mathbf{i}+4\mathbf{j}$ Hence, The vector $\mathbf{u}=10\mathbf{i}+4\mathbf{j}$ is orthogonal to $\mathbf{v}=-2\mathbf{i}+5\mathbf{j}$.