Question about vector equations of lines and planes

Question

Find the equation of the line going through the point $(2,-3,4)$ ,and which is perpendicular to the plane $ x+2y + 2z = 13$

So I tried this: the normal of the plane is $(1,2,2)$, random point on the line is $(x,y,z)$, so direction vector is $(x-2,y+3,z-4)$. We'll require $(x-2,y+3,z-4)\cdot (1,2,2) = 0$. This yields $x-2y+2z -4 =0$

let us take $z=y=1$. that makes $x=0$ and so $(0,1,1)$ is a random point on the plane; and so we get direction vector $(-2,4,-3)$. This gives a line equation of: $(2,-3,4) + t(-2,4,-3)$ ;

general point on such a line is: $(2-2t,4t-3,-3t+4)$. We also need to find it's point of intersection with the plane $ x+2y + 2z = 13$, so we substitute values of the general point to the plane $ x+2y + 2z = 13$. this however yields $0=9$ which is a little unsatisfying.

Answer 1

The reason why you get $0=9$ is that there is no intersection; the intersection is empty. Notice that the plane $P: x+2y+2z=13$ is parallel to the original plane, and so is the line. This means that either the line lies on the plane $P$ , or the line is parallel to it--and the second choice here is the case. If you start with a wrong assumption you may not get a good result (i.e., you are not guaranteed to, which you would be if you started with a true premise and argued correctly) , even if you argue correctly .

EDIT: Since the point lies on a line perpendicular to the plane thru $(2, -3,4)$, the line must be of the type : $L: (2,-3,4)+ t(1,2,2)=(2+t, -3+2t, 4+2t)=(x,y,z)$ . We then substitute this form into $x+2y+2z=13$, and find that $t=1$ is the solution ( makes sense, since both must intersect at a single point by a (co)dimension argument ). From $t=1$, we get that the intersection point is $(2+1, -3+2, 4+2)=(3,-1,6)$

Answer 2

The equation of the line $l$ which is passing through the point $P_0=(x_0,y_0,z_0)$ and along the non-zero vector $\vec u=\langle u_1,u_2,u_3\rangle$: $$\begin{align}&x=u_1t+x_0,\\ &y=u_2t+y_0,\\&z=u_3t+z_0,\,\,\text{ where $t\in\Bbb R$.} \end{align}\tag1$$ The equation can be written like $\color{purple}{\vec P=\vec u\,t+\vec P_0}$ for short. Not also that $\vec u$ is called a direction vector of the line.

The equation of the plane $H$ which is passing through the point $P_0=(x_0,y_0,z_0)$ and perpendicular to the non-zero vector $\vec n=\langle n_1,n_2,n_3\rangle$: $$n_1x+n_2y+n_3z=n_1x_0+n_2y_0+n_3z_0\tag2$$ The equation can be written like $\color{purple}{\vec n\cdot\vec P=\vec n\cdot\vec P_0}$ for short where $\cdot$ is the standart inner product of $\Bbb R^3$. Note also that $\vec n$ is called a normal vector of the plane.

Lemma: If $l\perp H$ then $\vec u=k\vec n$ for some non-zero $k\in\Bbb R$.

We are ready to solve the question. By $(2)$, a normal vector of the plane $H$, $$x+2y+2z=1,\tag3$$ is $\vec n=\langle1,2,2\rangle$. Taking $k=1$ in the Lemma, a direction vector of a line $l$ which is perpendicular to $H$ is $\vec u=\langle1,2,2\rangle$. Hence, since $l$ is passing through $(2,-3,4)$, by $(1)$, the equation of the line is $$(x,y,z)=(t+2,2t-3,2t+4)\tag4.$$ Solving $(3)$ and $(4)$ together we also have $t=1$ and $l\cap H=\{(3,-1,6)\}.$