# Find the equation of a plane given point, parallel line and angle between plane and line

Find the equation of a plane that crosses point $$P(-1,2,1)$$, that is parallel to the line $$p: x=0,y=-z$$ and its angle with line $$q: x=y, z=0$$ is $$\frac{\pi}{4}$$.

First lets write those two lines in canonical form. Lets observe normal vectors of two $$p$$ planes: $$\vec{n_{p_1}}=(1,0,0)$$ and $$\vec{n_{p_2}}=(0,1,1)$$. Their cross product will give direction vector of $$p$$. So, $$\vec{n_{p_1}}\times\vec{n_{p_2}}=(0,-1,1).$$ Lets take a point now thats on $$p$$ and satisfies both planes of $$p$$, e.g.: $$A(0,2,-2)$$. Now we have the canonical form of $$p$$: $$\frac{x-0}{0}=\frac{y-2}{-1}=\frac{z+2}{1}$$.

Same for $$q$$: $$\vec{n_{q_1}}=(1,-1,0)$$ and $$\vec{n_{q_2}}=(0,0,1)$$ and their cross product: $$\vec{n_{q_1}}\times\vec{n_{q_2}}=(-1,-1,0).$$ Lets take a point that satisfies both planes of $$q$$, e.g.: $$B(1,1,0)$$. Now we have the canonical form for $$q$$: $$\frac{x-1}{-1}\frac{y-1}{-1}\frac{z-0}{0}$$.

$$\sin\alpha=\frac{\vec{n_\pi}\vec{q}}{|\vec{n_\pi}||\vec{q}|}$$ where $$\vec{n_\pi}$$ is normal to the plane we're looking for. $$\alpha=\frac{\pi}{4}$$ so $$\frac{\vec{n_\pi}\vec{q}}{|\vec{n_\pi}||\vec{q}|}=\frac{\sqrt{2}}{2}$$. If $$(k,l,m)$$ is normal vector to the plane we're looking for then it's perpendicular to the direction vector of $$p$$ because we know that the line and plane are parallel: $$\vec{p}(k,l,m)=0$$ -> $$(0,-1,1)(k,l,m)=0$$ -> $$-l+m=0$$ -> $$m=l$$. Now I have another equation from the angle condition.

$$\frac{(k,l,m)(-1,-1,0)}{\sqrt{k^2+l^2+m^2}\sqrt{2}}=\frac{\sqrt{2}}{2}$$.

This is where I'm stuck. If I somehow manage to get normal vector, I have the point $$P(-1,2,1)$$ and with the normal vector I have the equation of my plane. What should I do next? Any tips would be appreciated!

• Not sure what;s wrong with $\frac{(k,m,m)\cdot(-1,-1,0)}{\sqrt{k^2+2m^2}}=1$. I'd square both parts and let $x=\frac{m}{k}$: $(1+x)^2=1+2x^2$ hence we have $x=0$ or $x=2$ so $(k,0,0)$ and $(k,2k,2k)$ are solutions for the desired vector. Commented Jul 14, 2020 at 13:20
• @AlexeyBurdin So now I have vector k(1,2,2) where I can put k=1 and use (1,2,2) as my normal vector through the point P ? Simple as that? Commented Jul 14, 2020 at 13:32
• You can verify if the conditions hold in case of a doubt. Yes, that simple.) Don't forget of $(1,0,0)$ too. Commented Jul 14, 2020 at 13:36
• But I have a restriction for that $m \neq 0$ so I observe the case where m=2k. Right? Commented Jul 14, 2020 at 13:40
• I don't see from where $m\ne 0$ follows. Doesn't $(1,0,0)$ fit? Another question why we can divide by $k$: if $k=0$ then $\frac{(k,m,m)\cdot(-1,-1,0)}{\sqrt{k^2+2m^2}}=\frac{-m}{\sqrt{2}|m|}\ne 1$. Commented Jul 14, 2020 at 14:16

Let $$\hat{n} = (n_1,n_2,n_3)$$ be the normal vector of the plane with $$|\hat{n}| = 1$$ and let the equation of the plane be $$n_1x+ n_2y+n_3z+D=0$$ for some $$D\in\Bbb{R}$$.
Line $$p$$ is the set of points $$\alpha(0,1,-1)$$ for $$\alpha\in\Bbb{R}$$. It is parallel to the plane so $$(0,1,-1) \perp \hat{n}$$, which implies $$0 = (0,1,-1) \cdot \hat{n} = n_2-n_3.$$ On the other hand, line $$q$$ is the set of points $$\alpha(1,1,0)$$ for $$\alpha\in\Bbb{R}$$. It punctures the plane at an angle $$\frac\pi4$$ which means that the angle of $$(1,1,0)$$ and $$\hat{n}$$ is the complementary angle which is again $$\frac\pi4$$. Therefore $$\frac{\sqrt{2}}2 = \cos \measuredangle((1,1,0),\hat{n}) = \frac{(1,1,0)\cdot \hat{n}}{|(1,1,0)|\cdot|\hat{n}|} = \frac{n_1+n_2}{\sqrt2}$$ so $$n_1+n_2=1$$. Now combine equations $$\begin{cases} n_1+n_2=1\\ n_2-n_3=0\\ n_1^2+n_2^2+n_3^2=0\\ \end{cases}$$ to obtain two solutions for $$\hat{n}$$: $$\hat{n} = \frac13(1,2,2), \quad \hat{n}=(1,0,0).$$ Finally use that $$P = (-1,2,1)$$ is contained in the plane so $$(-1,2,1)\cdot\hat{n}+D = 0$$ which gives you $$D$$. The resulting two planes are $$x+2y+2x-5=0, \quad x+1=0.$$