An implicit equation for the plane passing through the points (5, 2, -1), (4, 4, 2), and (4, 2, -3) is -4x+1y+2z=0

An implicit equation for the plane passing through the points (5, 2, -1), (4, 4, 2), and (4, 2, -3) is -4x+1y+2z=0

The Correct Answer and Explanation is:

ChatGPT said:

To find the implicit equation of the plane passing through the points (5,2,−1)(5, 2, -1)(5,2,−1), (4,4,2)(4, 4, 2)(4,4,2), and (4,2,−3)(4, 2, -3)(4,2,−3), we can follow these steps:

Step 1: Find two vectors that lie on the plane

We start by creating two vectors from the given points. Let’s use A=(5,2,−1)\mathbf{A} = (5, 2, -1)A=(5,2,−1), B=(4,4,2)\mathbf{B} = (4, 4, 2)B=(4,4,2), and C=(4,2,−3)\mathbf{C} = (4, 2, -3)C=(4,2,−3).

1. Vector AB\mathbf{AB}AB is the vector from AAA to BBB: AB=B−A=(4−5,4−2,2−(−1))=(−1,2,3)\mathbf{AB} = B – A = (4 – 5, 4 – 2, 2 – (-1)) = (-1, 2, 3)AB=B−A=(4−5,4−2,2−(−1))=(−1,2,3)
2. Vector AC\mathbf{AC}AC is the vector from AAA to CCC: AC=C−A=(4−5,2−2,−3−(−1))=(−1,0,−2)\mathbf{AC} = C – A = (4 – 5, 2 – 2, -3 – (-1)) = (-1, 0, -2)AC=C−A=(4−5,2−2,−3−(−1))=(−1,0,−2)

Step 2: Find the normal vector to the plane

To find the normal vector n\mathbf{n}n to the plane, we take the cross product of vectors AB\mathbf{AB}AB and AC\mathbf{AC}AC:n=AB×AC\mathbf{n} = \mathbf{AB} \times \mathbf{AC}n=AB×AC

We compute the cross product:n=∣ijk−123−10−2∣\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 3 \\ -1 & 0 & -2 \end{vmatrix}n=​i−1−1​j20​k3−2​​

Expanding the determinant:n=i∣230−2∣−j∣−13−1−2∣+k∣−12−10∣\mathbf{n} = \mathbf{i} \begin{vmatrix} 2 & 3 \\ 0 & -2 \end{vmatrix} – \mathbf{j} \begin{vmatrix} -1 & 3 \\ -1 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -1 & 2 \\ -1 & 0 \end{vmatrix}n=i​20​3−2​​−j​−1−1​3−2​​+k​−1−1​20​​n=i(2(−2)−3(0))−j(−1(−2)−3(−1))+k(−1(0)−2(−1))\mathbf{n} = \mathbf{i} (2(-2) – 3(0)) – \mathbf{j} (-1(-2) – 3(-1)) + \mathbf{k} (-1(0) – 2(-1))n=i(2(−2)−3(0))−j(−1(−2)−3(−1))+k(−1(0)−2(−1))n=i(−4)−j(2+3)+k(0+2)\mathbf{n} = \mathbf{i}(-4) – \mathbf{j}(2 + 3) + \mathbf{k}(0 + 2)n=i(−4)−j(2+3)+k(0+2)n=−4i−5j+2k\mathbf{n} = -4\mathbf{i} – 5\mathbf{j} + 2\mathbf{k}n=−4i−5j+2k

Thus, the normal vector is n=(−4,−5,2)\mathbf{n} = (-4, -5, 2)n=(−4,−5,2).

Step 3: Write the equation of the plane

The equation of the plane is given by the dot product of the normal vector and the vector from any point on the plane to a general point (x,y,z)(x, y, z)(x,y,z):n⋅(x−x1,y−y1,z−z1)=0\mathbf{n} \cdot (x – x_1, y – y_1, z – z_1) = 0n⋅(x−x1​,y−y1​,z−z1​)=0

Using point A(5,2,−1)A(5, 2, -1)A(5,2,−1), we get:−4(x−5)−5(y−2)+2(z+1)=0-4(x – 5) – 5(y – 2) + 2(z + 1) = 0−4(x−5)−5(y−2)+2(z+1)=0

Expanding:−4x+20−5y+10+2z+2=0-4x + 20 – 5y + 10 + 2z + 2 = 0−4x+20−5y+10+2z+2=0−4x−5y+2z+32=0-4x – 5y + 2z + 32 = 0−4x−5y+2z+32=0

So the implicit equation of the plane is:−4x−5y+2z=−32-4x – 5y + 2z = -32−4x−5y+2z=−32

or equivalently:4x+5y−2z=324x + 5y – 2z = 324x+5y−2z=32

Conclusion:

The equation −4x+y+2z=0-4x + y + 2z = 0−4x+y+2z=0 does not match the correct result from the steps above. The correct equation of the plane passing through the points (5,2,−1)(5, 2, -1)(5,2,−1), (4,4,2)(4, 4, 2)(4,4,2), and (4,2,−3)(4, 2, -3)(4,2,−3) is:4x+5y−2z=324x + 5y – 2z = 324x+5y−2z=32