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Problem 581

# Find the equation of the plane through the points $\mathrm{P}_{1}(1,2,-1), \mathrm{P}_{2}(2,3,1)$$, and$$\mathrm{P}_{3}(3,-1,2)$

Expert verified
The equation of the plane through the points $$P_1(1,2,-1), P_2(2,3,1)$$, and $$P_3(3,-1,2)$$ is $$9x + y - 5z - 16 = 0$$.
See the step by step solution

## Step 1: Find two directional vectors on the plane

We have the points P1(1, 2, -1), P2(2, 3, 1), and P3(3, -1, 2). We can find two directional vectors on the plane by subtracting coordinates of the points as follows: Vector A = P2 - P1 = (2-1, 3-2, 1-(-1)) = (1, 1, 2) Vector B = P3 - P1 = (3-1, -1-2, 2-(-1)) = (2, -3, 3)

## Step 2: Find the normal vector to the plane

To find the normal vector N, we can compute the cross product of the two directional vectors A and B. The cross product is given by: N = A x B = $$\begin{pmatrix} i & j & k \\ 1 & 1 & 2 \\ 2 & -3 & 3 \end{pmatrix}$$ = i($$1(3) - 2(-3)$$) - j($$1(3) - 2(2)$$) + k($$1(-3) - 2(1)$$) = i(3 + 6) - j(3 - 4) + k(-3 - 2) = 9i + j - 5k Thus, the normal vector N = (9, 1, -5).

## Step 3: Write the equation of the plane using the point-normal form

Now that we have a normal vector N = (9, 1, -5), we can use the point-normal form of the plane equation, which is given by: N . (X - P1) = 0 where "." denotes the dot product, X = (x, y, z) is any point on the plane, and P1 is one of the given points on the plane, say (1, 2, -1). Substituting the values of N and P1 into the equation, we get: (9, 1, -5) . (x-1, y-2, z-(-1)) = 0 Calculating the dot product, we get: 9(x-1) + 1(y-2) - 5(z+1) = 0

## Step 4: Simplify the equation

Finally, we simplify the plane equation: 9x - 9 + y - 2 - 5z - 5 = 0 9x + y - 5z - 16 = 0 Hence, the equation of the plane through the points P1, P2, and P3 is: $$9x + y - 5z - 16 = 0$$

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