Simple Line of intersection between two planes $x + 3y = 7$ and $2y + z = 4$

Question

Let L be the line of intersection between the planes x + 3y = 7 and 2y + z = 4, and let P be the point (3, 1, 1).

Determine the equation (in normal form) of the plane containing L and P.

I struggle with getting vectors from the two planes to multiply, can anyone show me how?

$L =(1, 2, 0)+t(3, −1, 2)$. But i somehow fail to see how my book got that.

Answer 1

To obtain $L$, parameterize it with $y=2+t$ and, from the two given planes, we have $x=1-3t$ and $z=-2t$, i.e.

$$L= (1,2,0)+t(-3,1,-2)$$

with $Q(1,2,0)$ and $\vec d =(-3,1,-2)$. Let $A(x,y,z)$ be any point on the plane and $P(3,1,1)$. Then, the normal vector of plane is

$$\vec n=\vec{QA}\times \vec{QP}= (x-1,y-2,z)\times (2,-1,1)\\ =(-2+y+z,1-x+2z,5-x-2y)$$

which satisfies $\vec n \cdot \vec d = 0$, i.e.

$$ (-2+y+z,1-x+2z,5-x-2y)\cdot (-3,1,-2)=0$$

which yields the equation of the plane

$$x+y-z=3$$

Answer 2

Hint:

A plane through the intersection line of the planes belongs to the pencil of planes: $$(x+3y)+\lambda(2y+z)=7+4\lambda\qquad(\lambda\in\mathbf R).$$ Determine the value of $\lambda$ such that the corresponding plane passes though $P$.

Answer 3

Let $\vec{l}(a,b,c)$ be a vector parallel to the line.

Thus, $$a+3b=0$$ and $$2b+c=0,$$ which gives $$c=-2b,$$ $$a=-3b$$ and for $b=-1$ we obtain: $$\vec{l}(3,-1,2)$$ and since a point $(7,0,4)$ is placed on the line we got an equation of this line: $$(x,y,z)=(7,0,4)+t(3,-1,2).$$

Now, let $\vec{n}(a,b,c)$ be a normal to the needed plane.

Thus, $$3a-b+2c=0$$ and since a vector $$(7-3,0-1,4-1)$$ or $$(4,-1,3)$$ is parallel to the needed plain, we obtain also $$4a-b+3c=0.$$ Thus, $a=-c$, $b=-c$ and $$\vec{n}(1,1,-1),$$ which gives the following equation. $$(x-3)+(y-1)-(z-1)=0$$ or $$x+y-z-3=0.$$