Find the equation of the plane that passes through the line of intersection of two planes and a point

Find the equation of the plane that passes through the line of intersection of the planes $$2x-3y-z +1 =0$$ and $$3x+5y-4z+2=0$$, and that also passes through the point $$(3,-1,2)$$

$$\vec n_1 = [2,-3,-1]$$

$$\vec n_2 = [3,5,-4]$$

$$\vec n_1 \times \vec n_2 = [17,5,19]$$

$$[17,5,19]$$ is the direction vector of the line of intersection.

now,

$$[17,5,19] \cdot \vec n_3 = 0 \\ [17,5,19] \cdot [a,b,c] =0 \\ 17a+5b+19c = 0$$

Let $$a =1, b=1$$ $$17+5 + 19c =0 \\ 19c = -22 \\ c=-{22 \over 9}$$

$$\vec n_3 = [1,1,-{22\over 19}] \\ \equiv [19,19,-22]$$

So the scalar equation is $$19x +19y- 22z+ D= 0$$

Substitute $$(3,-1,2)$$

$$19(3) +19(-1) -22(2) +D = 0 \\ D = 6$$

$$19x -19y -22z + 6 =0$$ is the equation of the plane.

However the answer says it is $$14x +17y -17z +9 = 0$$

• use family of planes – maveric Nov 16 '18 at 18:09

Another hint:

Equations of all planes passing through the intersection of planes 2x-3y-z+1 = 0 and 3x+5y-4z+2=0 have a shape

$$\alpha (2x-3y-z+1) + \beta (3x+5y-4z+2)=0$$

The alpha and beta constants are not at the same time equal to 0. Suppose that alpha $$\neq 0$$ and divide:

$$\frac{\beta}{\alpha}=\gamma \quad \Rightarrow \quad 2x-3y-z+1 + \gamma (3x+5y-4z+2)=0$$

The plane passes through the point $$(3,-1,2) \Rightarrow \gamma = 4$$ and the equation of the search plane will be

$$x + 17 y - 17 z + 9 = 0$$

Find two points $$P$$ and $$Q$$ on the line of intersection of the planes $$2x−3y−z+1=0$$ and $$3x+5y−4z+2=0.$$

Then with $$A=(3,−1,2)$$ construct vectors $$AP$$ and $$AQ$$

The normal vector to your plane is the common perpendicular of $$AP$$ and $$AQ$$

Having a point and the normal vector you can easily find the equation of the plane.