Finding the value of $k$ for parallel/orthogonal planes

Question

Given two planes

$\pi_1: (1+k)x-3y+6z-4=0$

$\pi_2: x+(5+k)y+3kz+1=0$

Find the value of $k$ such that $\pi_1$ is parallel to $\pi_2$ and when $\pi_1$ is perpendicular to $\pi_2$

Answer 1

HINT: two planes are parallel/orthogonal iff their normal vectors are parallel/orthogonal.

Let's show how to solve the exercise using the hint:

As @Meadara said, from the plane equation $$ax+by+cz=d$$ one easily reads out the normal vector $(a,b,c)$, so the normal vectors are $\vec{n}_1=(1+k,-3,6)$ and $\vec{n}_2=(1,5+k,3k)$. By the hint the two planes are parallel/orthogonal if the two normal vectors are parallel/orthogonal.

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But two vectors $\vec{v},\vec{w}$ are parallel iff $\vec{v}=\lambda\vec{w}$ for some non zero constant $\lambda$. In our situation this translates to $$\begin{pmatrix}1 \\ 5+k \\ 3k \end{pmatrix}=\lambda \begin{pmatrix}1+k \\ -3 \\ 6 \end{pmatrix}=\begin{pmatrix}\lambda (1+k) \\ -3\lambda \\ 6\lambda \end{pmatrix}.$$ The last entry gives the condition $k=2\lambda$, the second $5+k=-3\lambda$, so the first into the second gives $$5+2\lambda=-3\lambda\iff\lambda=-1\Rightarrow k=-2,$$ and plugging these into the first coordinate we can check that indeed they satisfy also $1=\lambda(1+k)$ and so the planes are parallel iff $\lambda=-1\Rightarrow k=-2$.

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Orthogonality is easier : two vectors are orthogonal iff their scalar product is zero. The scalar product of $\vec{n_1}$ and $\vec{n_2}$ is $$\vec{n_1}\cdot \vec{n_2}=\begin{pmatrix}1 \\ 5+k \\ 3k \end{pmatrix}\cdot \begin{pmatrix}1+k \\ -3 \\ 6 \end{pmatrix}=1+k-3(5+k)+18k=16k-14$$ which is zero iff $k=\frac{14}{16}=\frac{7}{8}$.