Finding a line that is parallel to two planes in 3D

I have a question asking me to find a line in the symmetrical equation that passes through $$(1,1,1)$$ and also parallel to the following two planes

• $$\pi_1 = [x,y,z] = [1,-3,4]+r[-1,3,2]+s[2,1,2]$$
• $$\pi_2 = [x,y,z] = [1,2,1]+r[1,-3,2]+s[1,1,2]$$

I manage to find the normal equation of these two planes:

• $$\pi_1 = 4x+6y-7z=-42$$
• $$\pi_2 = -8x+4z=-4$$

Then I am not sure how to proceed with this question. However, we can see that these two planes are not even parallel to each other, so how can we find a line that is parallel two both of them?

The answer key provided the solution as: $$\frac{x-1}{3}=\frac{y-1}{5}=\frac{z-1}{6}$$, but I drew the line in Geogebra and I do not think the line is parallel to both planes.

• Are you sure you calculated the correct normal form of the plane $\pi_2$? Nov 28, 2019 at 20:44
• Hmmm but I just tried with a calculator, it holds $4*1+6*-3-7*4 = -42$ Nov 28, 2019 at 20:48
• Yes sorry, indeed it is correct!
– user
Nov 28, 2019 at 20:49
• Shouldn't it be $\pi_2 : -8x+4z=-4$? Nov 28, 2019 at 20:50
• Oh right, sorry for the mistake I made. I gonna change it now Nov 28, 2019 at 20:53

To be parallel to both planes, the direction vector $$\vec v$$ of the line must be orthogonal to both normal vectors, then a direction vectors is given by
$$\vec v=\vec n_1\times \vec n_2=\begin{vmatrix}\hat i&\hat j&\hat k\\4&6&-7\\-8&0&-4\end{vmatrix}$$
• By cross product we obtain $(-24,-40,48)$ that is a multiple of $(3,5,-6)$. Therefore it seems there is a typo in the book. I think that using cross product is a very effective way in this case.