I am new to working with vectors, and I cannot understand the textbook's explanation of how to find perpendicular vectors, so I was hoping y'all could help me!
The problem reads:
Let $v = (2, 3)$. Suppose $w \in \mathbb{R}^2$ is perpendicular to $v$, and that $\|w\| = $5. This determines $w$ up to sign. Find one such $w$.
I found how to do parallel vectors from here, and have tried using it to get $\left(\frac{10}{\sqrt{13}},\frac{15}{\sqrt{13}}\right)$, then multiplying the second term by $-1$ so I would satisfy the algebraic definition of "perpendicular". However, the program is flagging my answer as wrong.
I also tried this formula: $v\cdot w=\|v\|\cdot\|w\|\cos(\theta)$, where I'm assuming I can rewrite it as: $$w=\frac{\|v\|\cdot\|w\|\cos(\theta)}{v},$$ and then I'd get the answer, but the program isn't accepting it either.
Thanks so much!
Edit: The answer I get from the second part (and pardon me if I'm doing this really wrong), is $\left(\frac{5\sqrt{13}}{2},\frac{5\sqrt{13}}{3}\right).$
Best Answer
A vector $(a,b)$ is perpendicular to the vectors $(b,-a)$ and $(-b,a)$ since the dot products are 0. One says that $(b,-a)$ and $(-b,a)$ are orthogonal or normal to vector $(a,b)$.
\begin{eqnarray} (a,b)\cdot(b,-a)&=&ab-ba=0\\ (a,b)\cdot(-b,a)&=&-ab+ba=0 \end{eqnarray}
One can find a unit normal vector to $(a,b)$ by dividing any normal vector by the magnitude $\Vert(a,b)\Vert=\sqrt{a^2+b^2}$. One may multiply a unit normal vector by any non-zero scalar to obtain a normal vector of any length.
So you begin with a normal vector to $(2,3)$ such as $(-3,2)$, divide by the magnitude $\sqrt{13}$ and multiply by $5$ to get one of the two possible answers
$$ \left(-\frac{15}{\sqrt{13}},\frac{10}{\sqrt{13}}\right)$$