# Some questions involving two lines and two planes in $\mathbb{R}^3$

Let $$d_1:\begin{cases} 2x+y-z=1 \\ x-z=2\end{cases}$$ and $$d_2:\begin{cases} x-y+2z=1 \\ x-y=2\end{cases}$$ be two lines in $$\mathbb{R}^3$$.

I have to find the plane $$\pi_1$$ that contains $$d_1$$ and is parallel to $$d_2$$ and the plane $$\pi_2$$ that is perpendicular on $$\pi_1$$ and contains $$d_2$$.

Here's what I did. The direction of $$d_1$$ is given by the cross product of the normal vectors of the two planes whose intersection it is. So, the direction of $$d_1$$ is the vector $$\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -1 \\ 1 & 0 & -1 \end{vmatrix}=(-1, 1, -1)$$. Similarly, the direction of $$d_2$$ is $$\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -1 & 2 \\ 1 & -1 & 0 \end{vmatrix}=2(1, 1, 0)$$, so we may consider the direction to be $$(1, 1, 0)$$ too.\

Now, $$(1, -2, -1)\in d_1$$, so $$\pi_1:\begin{vmatrix} x-1 & -1 & 1 \\ y+2 & 1 & 1 \\ z+1 & -1 & 0\end{vmatrix}=0$$. We get that $$\pi_1 : x-y-2z=5$$.

Since $$\pi_2 \perp \pi_1$$, their normal vectors must be perpendicular. We may take the normal vector of $$\pi_2$$ to be $$(1, 1, 0)$$ and now since $$(2, 0, -\frac{1}{2})\in d_2$$ we get that $$\pi_2: 1(x-2)+1(y-0)+0(z+\frac{1}{2})=0$$, which means that $$\pi_2 :x+y=2$$.

I think that this is correct, but one of the subsequent questions implied that $$d_1$$ and $$\pi_2$$ intersect and this is false with the relations I have. Did I do anything wrong?

You found the equation of $$\pi_1$$ correctly but for $$\pi_2$$, please note that it is perpendicular to $$\pi_1$$ but also contains line $$d_2$$.
Normal vector to $$\pi_1$$ is $$(1, -1, -2)$$. Also, direction vector of line $$d_2$$ is $$(1, 1, 0)$$. Normal vector to $$\pi_2$$ must be perpendicular to both vectors.
So the normal vector to $$\pi_2$$ can be obtained using cross product of $$(1, -1, -2)$$ and $$(1, 1, 0)$$ which gives us direction $$(1, -1, 1)$$.
As point $$(2, 0, -\dfrac{1}{2})$$ is on line $$d_2$$ (and so also on the plane $$\pi_2$$), equation of $$\pi_2$$ is $$2x-2y+2z = 3$$.
• Ahh, I see. My $\pi_2$ doesn't actually contain the line $d_2$, right? Jun 19, 2021 at 17:19
• No, it does not contain line $d_2$. In fact, the normal vector of $\pi_2$ in your working has the same direction as line $d_2$, in other words, it is a plane that is perpendicular to $d_2$. It intersects the line at point $(2, 0, -1/2)$. Jun 19, 2021 at 17:25