# A Plane through 3 points

1. Oct 24, 2008

### roam

This is a very simple problem but I'm confused because the book's answer is completely different from mine;

Find the parametric vector equation of the plane that passes through the points P(3,1,2), Q(1,3,2) and R(1,1,1).

My attempt at a solution

This is simple, the parametric equations has to be: OX = OP + λPQ + μPR, λ,μ $$\in$$ R

$$\left(\begin{array}{ccc}3\\1\\2\end{ar ray}\right)$$ + $$\lambda$$ $$\left(\begin{array}{ccc}-2\\2\\0\end{ar ray}\right)$$ + $$\mu$$ $$\left(\begin{array}{ccc}-2\\0\\-1\end{ar ray}\right)$$

I'm confident that this is the right answer & here's a diagram of a similar situation.

But the correct answer according to the book has to be:

$$\lambda$$ $$\left(\begin{array}{ccc}3\\1\\2\end{ar ray}\right)$$ + $$\mu$$ $$\left(\begin{array}{ccc}1\\3\\2\end{ar ray}\right)$$ + $$(1- \lambda -\mu)$$ $$\left(\begin{array}{ccc}1\\1\\1\end{ar ray}\right)$$

So, what's wrong?

Last edited: Oct 24, 2008
2. Oct 24, 2008

### Dick

There's nothing wrong. There's an infinite number of ways to parametrize the plane. Both answers are right. If you look at the book answer and combine the lambdas and mus, you'll see that it's OR+lambda*PR+mu*QR. Same thing.

3. Oct 24, 2008

### Staff: Mentor

I don't see anything wrong. Your "equation" (there's not actually an equals in what you have) is satisfied by each of the three given points when 1) lambda = 0, mu = 0; 2) lambda = 1, mu = 0; 3) lambda = 0, mu=1.

The book's "equation" is also satisfied by various combinations of lambda and mu, each with a value of 0 or 1.

4. Oct 26, 2008

### roam

Ok thank you. I also have another question:

Find a normal vector to the plane through the points P(2, 0, 2), Q(3, 1, 4) and R(1, 3, 2).

My Attempt:

Parametric vector equation of the plane is;

$$\left(\begin{array}{ccc}2\\0\\2\end{ar ray}\right)$$ + $$\lambda$$ $$\left(\begin{array}{ccc}1\\1\\2\end{ar ray}\right)$$ + $$\mu$$ $$\left(\begin{array}{ccc}-1\\3\\0\end{ar ray}\right)$$

The vectors PQ and PR are two linearly independent direction vectors for the plane. Hence according to a worked example in my book, PQ × PR is a normal vector to the plane. Now the coordinates of PQ = OQ − OP are: (1, 1, 2) and the coordinates of PR = OR − OP are: ( − 1, 3, 0).

(1, 1, 2) × ( − 1, 3, 0) = (-6,4,0)

But this is incorrect. The corrcet answer is that the coordinates of a normal vector are: ( − 6, − 2, 4). What did I do wrong here?
Thanks.

5. Oct 26, 2008

### HallsofIvy

Staff Emeritus
What you did was calculate the cross product incorrectly. How did you get (-6, 4, 0)?

6. Oct 26, 2008

### roam

Hi! Actually I made an error in my previous post. The cross product is defined by;

u × v = $$\left(\begin{array}{ccc}i&j&k\\u1&u2&u3\\v1&v2&v3\end{ar ray}\right)$$ = $$\left(\begin{array}{ccc}u2&u3\\v2&v3\end{ar ray}\right)$$ i - $$\left(\begin{array}{ccc}\\u1&u3\\v1&v3\end{ar ray}\right)$$ j + $$\left(\begin{array}{ccc}\\u1&u2\\v1&v2\end{ar ray}\right)$$ k

So, PQ × PR is;

$$\left(\begin{array}{ccc}i&j&k\\1&1&2\\-1&3&0\end{ar ray}\right)$$ = i $$\left(\begin{array}{ccc}1&2\\3&0\end{ar ray}\right)$$ - j $$\left(\begin{array}{ccc}1&2\\-1&0\end{ar ray}\right)$$ + k $$\left(\begin{array}{ccc}1&1\\-1&3\end{ar ray}\right)$$

The coordinates of a normal vector are: = (6, -2, 4)

Now, I don't understand why the right answer is supposed to be ( − 6, − 2, 4)! I used the formula so why did I get the wrong answer? How come they got a -6?

7. Oct 26, 2008

### gabbagabbahey

$\begin{vmatrix} 1 &2 \\ 3 &0 \end{vmatrix}=(1)(0)-(2)(3)=-6$

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