Problem on finding a perpendicular vector
$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
Find a 2d vector that is perpendicular to $\langle 2,3 \rangle$. Verify that it is perpendicular.
Solution
Recall that
Finding a perpendicular vector
A vector perpendicular to $\langle a,b \rangle$ is $\langle -b, a \rangle$.
Hence, a vector perpendicular to $\langle 2, 3 \rangle$ is $\langle -3, 2 \rangle$.
To verify these are perpendicular, we recall
Dot product of perpendicular vectors is zero
If $\mathbf{x}$ and $\mathbf{y}$ are perpendicular if and only if $\mathbf{x} \cdot \mathbf{y} = 0$.
Direct computation of dot product
We can verify these vectors are perpendicular by computing
\begin{align}
\langle 2,3 \rangle \cdot \langle -3, 2\rangle &= 2 \cdot (-3) + 3 \cdot 2\\ &= 0
\end{align}
Related topics
Multivariable calculus
(147 problems)
Vectors
(55 problems)
Ways of finding a perpendicular vector
(6 problems)
Dot product
(41 problems)
Dot product is defined by $\langle x_1, x_2, x_3 \rangle \cdot \langle y_1, y_2, y_3 \rangle = x_1 y_1 + x_2 y_2 + x_3 y_3.$
(21 problems)
The vectors $\mathbf{x}$ and $\mathbf{y}$ are perpendicular if and only if $\mathbf{x} \cdot \mathbf{y} = 0$.
(13 problems)
