Plane that passes through the point (−3, 2, 1) and contains the line of intersection of the planes x + y − z = 4 and 4x − y + 5z = 2

Question

Find an equation of the plane. The plane that passes through the point (−3, 2, 1) and contains the line of intersection of the planes x + y − z = 4 4x − y + 5z = 2

I know the normal to plane 1 is <1,1,-1> and the normal to plane 2 is <4,-1,5>. The cross product of these 2 would give a vector that is in the plane I need to find.

P1 x P2 = <4,-9,-5>

So now I have a point (-3,2,1) and a vector <4,-9,-5> on the plane but I'm not sure what to do next.

Answer 1

The line passes $z=0$ at $(1.2, 2.8, 0)$.

A vector from this point to $(-3,2,1)$ is $(4.2,0.8,-1)$.

This vector cross $(4,-9,-5)$ is $(13,-17,41)$, which is the normal of the plane.

The result is:

$$13(x-(-3))+(-17)(y-2)+(41)(z-1)=0$$

This is the scalar product of a vector in the plane and the normal vector.

Answer 2

Let $\Pi$ be the plane that we seek. Setting $z = 0$ in the two equations and solving simultaneously for $x$ and $y$, we find that $(6/5,14/5,0)$ is a point in the line of intersection of the two planes. So $(6/5,14/5,0)$ lies on $\Pi$. Since $(-3,2,1)$ also lies on $\Pi$, the vector from $(-3,2,1)$ to $(6/5,14,5,0)$, i.e., $\vec{w} = \langle 21/5,4/5,-1\rangle$, lies on $\Pi$. So the vector $\vec{n} = \vec{v} \times \vec{w}$ is normal to $\Pi$. So the equation of the plane is $\vec{n}\cdot \langle x + 3, y - 2, z - 1\rangle = 0$. Now simplify to get it in the form $ax + by + cz = d$.

Answer 3

If $\vec{r}=(x,y,z)$ is the position of a point on the intersection, then $<\vec{r},\vec{n}_1>=4$ and $<\vec{r},\vec{n}_2>=2$. Where $\vec{n}_1=(1,1,-1)$ and $\vec{n}_2=(4,-1,5)$ are the the normals to the planes $x+y-z=4$ and $4x-y+5z=2$ respectively. So for any $\vec{r}=(x,y,z)$ in the line, we have $<\vec{r},\vec{n}_1+\lambda\vec{n}_2>=4+2\lambda$.In other words we can say that: $$(x+y-z)+\lambda(4x-y+5z)=4+2\lambda$$

As $P=(-3,2,1)$ belongs to the line we have: $$((-3)+2-1)+\lambda(4.(-3)-(2)+5.(1))=4+2\lambda$$ $$-2-9\lambda=4+2\lambda\Rightarrow \lambda=-\frac{6}{11}$$

So substituting this $\lambda$ on the equation we have our answer. $$(x+y-z)-\frac{6}{11}(4x-y+5z)=4+2(-\frac{6}{11})$$ $$11x+11y-11z-24x+6y-30z=32$$ $$-13x+17y-41z=32$$