Find an equation of the line of intersection of the two given planes.

$x+2y+3z=4$ and $-2x+y-4z=6$

The two planes $x+2y+3z=4$and $-2x+y-4z=6$

The goal is to demonstrate that lines intersect and to determine the equation of a plane containing two lines.

The normal vector to the plane $x+2y+3z=4$is ${\mathbf{N}}_1=⟨1,2,3⟩$and the normal vector to the plane $-2x+y-4z=6$is ${\mathbf{N}}_2=⟨-2,1,-4⟩$

The normal vectors are not parallel since they are not scalar multiples of each other. As a result, the planes must cross.

Each of the normal vectors must be orthogonal to the line of intersection. Use the cross-product to obtain the direction vector, $\mathbf{d}$for the line of intersection.

The direction vector $\mathbf{d}$is given by:

$$\begin{gathered} \mathbf{d}={\mathbf{N}}_1\times {\mathbf{N}}_2 \\ =⟨1,2,3⟩\times ⟨-2,1,-4⟩ \end{gathered}$$

$=\left|\begin{array}{c}ijk\\ 123\\ -21-4\end{array}\right|$

$=⟨-11,-2,5⟩$

To find the point of intersection, set $z=0$in each of the equations $x+2y+3z=4$and $-2x+y-4z=6$

The solution of the following equations

$$\begin{gathered} x+2y=4(\text{Set}z=0) \\ -2x+y=6(\text{Set}z=0) \end{gathered}$$

is $x=-\frac{8}{5}$and $y=\frac{14}{5}$

Therefore, the point that lies on the both planes is $\left(-\frac{8}{5},-\frac{14}{5},0\right)$

The line of intersection is:

Therefore, line of intersection is $\mathrm{r}\left(t\right)=\left(-\frac{8}{5}-11t,\frac{14}{5}-2t,5t\right)$