Find the point Q so that the segment PQ is normal to the plane.

Question

I am trying to find a point $Q$, so that the segment $\overline{PQ}$ is normal to the plane $x-2y+z=4$, while $P=(1,1,-1)$.

I get that I have to use the normal vector, which based on the coefficients of the plane equation is $(1,-2,1)$, however, I can't seem to proceed any further, the worst part is that I actually solved this before, I remember maybe using a vector product the first time, which I tried again as $(Q-P)xN=0$, solving it for a variable, putting it in the plane equation got me $0=4$, what for some reason, beside not being a helpful result, seems a bit wrong.

I can expand my calculations if requested, however, I am not really sure I am taking the best/right approach; so, am I terribly wrong? If so, what would be the best way to proceed in this scenario?

Answer 1

I assume that you mean that point $Q$ has to be on the plane. There are many possible ways to approach this. Projections are probably the fastest way to do this, but let's just go full-algebraic on this one.

Let $Q = (q_1, q_2, q_3)$. Then we want to know how many scalar multiples of the normal vector we are away from $P$. That is, find $\lambda \in \mathbb{R}$ such that $Q + \lambda N = P$. This gives $(q_1 + \lambda, q_2 - 2\lambda, q_3 + \lambda) = (1,1,-1)$. This can be expanded into 3 linear equations. The $4$th equation comes from the condition that $Q$ should be in the plane: $q_1 - 2q_2 + q_3 = 4$. Solving all $4$ linear equations simultaneously should give you your point $Q$.

EDIT: I didn't address your method it seems. Doing $(Q - P) \times N = \vec{0}$ should work. Let's try it here: \begin{align} (q_1 - 1, q_2 - 1, q_3 + 1) \times (1,-2,1) &= (0,0,0)\\ (q_1 - 1, q_2 - 1, 2q_2 - q_1 + 5) \times (1,-2,1) &= (0,0,0) \text{ (sub plane equation)}\\ \left|\begin{array}{c c c} i & j & k \\ q_1 - 1 & q_2 - 1 & 2q_2 - q_1 + 5 \\ 1 & -2 & 1 \end{array} \right| &= (0,0,0)\\ (5q_2 - 2q_1 + 9, 2q_2 - 2q_1 - 6, -2q_1 - q_2 + 3) &= (0,0,0) \end{align} Hence we have, \begin{align} 5q_2 - 2q_1 + 9 &= 0\\ q_2 - q_1 + 3 &= 0\\ - q_2 - 2q_1 + 3 &= 0 \end{align} Adding the second and third gives $-3q_1 + 6 = 0 \Rightarrow q_1 = 2$. Hence, $q_2 = 2 - 3 = -1$. For consistency, $5(-1) - 2(2) + 9 = -5 - 4 + 9 = 0$. Using the plane equation, we have which gives $q_3 = -(2) + 2(-1) + 4 = 0$. Finally, we have $Q = (2,-1,0)$. So surprise surprise, it looks like $\overline{PQ} = (1,-2,1)$ so that $\lambda = -1$ if you did it the first way I suggested.

EDIT 2: Thanks to amd, we can recognize that $Q = P = \lambda N$ for some $\lambda \in \mathbb{R}$, which we can put into the plane equation as: \begin{align} (P + \lambda N) \cdot N &= 4\\ P \cdot N + \lambda (N \cdot N) &= 4\\ (1,1,-1) \cdot (1,-2,1) + \lambda (1,-2,1) \cdot (1,-2,1) &= 4\\ -2 + 6 \lambda &= 4\\ \lambda &= 1 \end{align} Hence, $Q = (1,1,-1) + (1,-2,1) = (2,-1,0)$.

Answer 2

The method that you tried should have worked. Expanding $(P-Q)\times N = 0$ produces the system $$\begin{align} y+2z &= -1 \\ x-z &= 2 \\ 2x+y &= 3. \end{align}$$ It’s pretty obvious that the third equation is not independent of the first two (this will always be the case with systems of linear equations derived from a cross product like this), so we can drop it. Adding the equation of the plane to the system gives you three equations in three unknowns again, which I hope you know how to solve. Without seeing the details of your work, it’s hard to guess where you went wrong.

There’s a much simpler way to go about solving this problem, though. The point $Q$ must be of the form $P+\lambda N$, so just substitute that into the equation of the plane. This produces a single linear equation in $\lambda$. Solve it and substitute back into $P+\lambda N$.

Answer 3

For $P = (1,1,-1)$ and $x - 2y + z = 4$

The normal line where a, b and c are the x, y and z coefficients is:

$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ which results in

$(x - 1) = -\frac{(y - 1)}{2} = (z + 1)$