Find values of $k$ for which the lines $3x - ky = 5$ and $(k^2 - 2)x + 3y = 4$ are perpendicular

Question

Problem:
I'd like to find the values of $k$ for which the two lines $$3x - ky = 5\quad\text{and}\quad (k^2 - 2)x + 3y = 4$$ are perpendicular.

I have been trying to solve this problem for hours, and I haven't found anything similar on YouTube (or at least not in my native language). If someone could please help me, I would greatly appreciate it.

Edit:
I'll add what I've been trying. (Sorry if something is translated badly, my english isn't that good).

I tried to isolate $y$ from the other terms in the equation, resulting in: $$\frac3kx - \frac5k = y$$ $$-\frac{k^2 - 2}3\,x + \frac43 = y$$ Then, I solved for the slopes as $$m_1 = \frac3k\quad\text{and}\quad m_2 = -\frac{k^2 - 2}3$$ and proceeded to multiply them together to obtain the value of $k$ for them to be perpendicular: $$\frac3k\cdot\left(-\frac{k^2 - 2}3\right) = -1$$ This is where I got stuck. When solving the equation, I only obtain one value for $k$, when I should be getting more than one.

I applied distributive: $$\frac{3(k^2-2)}{3k} = -1$$ And now I don't know what to do. I tried to simplify, but it didn't work. I tried moving the denominator to the other side, so I could do Ruffini after bringing it back as $+ 3k$, but I couldn't find any roots.
Please help

Answer 1

Note that $\frac{3}{-k}\times \frac{k^2-2}{3}=-1$ for perpendicularity.

This gives $\frac{3k^2-6}{-3k}=-1$, which then gives

$k^2-2=k$.

The only numbers that are two less than their squares are

$k=-1$ and $k=2$.

Answer 2

Once you got

$(3/k) \cdot (-(k^2 - 2)/3) = -1$

you can reduce both threes

$\frac 1k \cdot (-(k^2 - 2)) = -1$

and then multiply both sides by $k$:

$-(k^2 - 2) = -k.$

This makes an equation

$k^2 -k - 2 = 0$

so

$k^2 - 2\cdot\frac 12\cdot k + \frac 14 = 2\frac 14$

$\left(k-\frac 12\right)^2 = \frac 94$

$k - \frac 12 = \pm \frac 32$

and, finally

$k = -1$ or $k = 2.$

Answer 3

A quick track towards the answer $\,k\in\{-1,2\}\,$ is available via Hesse normal forms $$\begin{pmatrix}3\\-k\end{pmatrix}\cdot\begin{pmatrix}x\\ y\end{pmatrix}\:=\:5\qquad\begin{pmatrix}k^2-2\\ 3\end{pmatrix}\cdot\begin{pmatrix}x\\ y\end{pmatrix}\:=\:4\,.$$ The corresponding lines are perpendicular iff the (non-normalised) normal vectors are perpendicular, hence $$0\:=\:\begin{pmatrix}3\\-k\end{pmatrix}\cdot\begin{pmatrix}k^2-2\\ 3\end{pmatrix}\:=\:3\left(k^2-2\right)-3k\:=\: 3(k+1)(k-2)\,.$$

Answer 4

The dot product of two perpendicular vectors is 0.

$$ \bar {n_1} \cdot \bar{n_2} = 0 $$

Dot product multiplication is as follows:

$$ A = \langle a_x, a_y \rangle $$ $$ B = \langle b_x, b_y \rangle $$ $$A \cdot B = a_xb_x + a_yb_y + a_zb_z$$

Our vectors:

$$ 3x-ky-5 = 0$$ $$(k^2-1)x +3y-4=0 $$

So our dot product is:

$$ \langle 3, -k \rangle \cdot\langle (k^2-1), 3\rangle$$

Set our dot plot equal to 0,

$$ 3(k^2-1) -3k = 0 $$

Finally solve for k