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Problem 588

# Show that the points $$\mathrm{P}=(0,2,0), \mathrm{Q}=(1,2,1)$$ and $$\mathrm{R}=(2,2,1)$$ are noncollinear. Then find the unique plane passing through these points.

Expert verified
The points P, Q, and R are noncollinear because their direction vectors (1, 0, 1) and (2, 0, 1) are not proportional. The unique plane passing through these points has the equation $$x + y - 2 = 0$$.
See the step by step solution

## Step 1: Check for collinearity

First, we need to form vectors based on the given points P, Q, and R. To do this, we'll use two pairs of points, such as PQ and PR: $\vec{PQ} = \vec{Q} - \vec{P} = (1, 2, 1) - (0, 2, 0) = (1, 0, 1)$ $\vec{PR} = \vec{R} - \vec{P} = (2, 2, 1) - (0, 2, 0) = (2, 0, 1)$ Now we will check if these vectors are proportional. For two vectors to be proportional, their components must be proportional as well: $\frac{1}{2} = \frac{0}{0} = \frac{1}{1}$ Since 0/0 is undefined and 1/2 is not equal to 1/1, the vectors PQ and PR are not proportional. Thus, points P, Q, and R are noncollinear.

## Step 2: Find the unique plane passing through the points

Now that we've established that the points P, Q, and R are noncollinear, we can find the unique plane passing through them. We will use the scalar equation of a plane in the form: $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$ To find values for A, B, and C, we can use the cross product of the direction vectors we found in Step 1, PQ and PR: $\vec{n} = \vec{PQ} \times \vec{PR} = \begin{bmatrix} i & j & k \\ 1 & 0 & 1 \\ 2 & 0 & 1 \end{bmatrix} = (-1)i -(1)j + (0)k = (-1, -1, 0)$ Here, $$\vec{n} = (-1, -1, 0)$$ is the normal vector of the plane. Thus, A = -1, B = -1, and C = 0. Choosing point P (0, 2, 0) as the reference point, our equation becomes: $-1(x - 0) -1(y - 2) + 0(z - 0) = 0$ Simplifying the equation, we get: $x + y - 2 = 0$ This is the unique plane passing through the points P, Q, and R.

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