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Question Find the plane equation ax+by+cz=0ax + by + cz = 0 that passes through the points (1,0,2),(−1,1,−2)(1,0,2), (-1,1,-2), and the origin.

Studdy Solution

STEP 1

Assumptions
1. The plane contains the points (1,0,2)(1,0,2), (−1,1,−2)(-1,1,-2), and the origin (0,0,0)(0,0,0).
2. The equation of the plane is in the form ax+by+cz=0ax + by + cz = 0.
3. We need to find the values of aa, bb, and cc.

STEP 2

To find the equation of the plane, we can use the fact that the normal vector to the plane can be found by taking the cross product of two vectors that lie in the plane.

STEP 3

First, we find two vectors that lie in the plane by subtracting the coordinates of the origin from the other two points.
Vector AB⃗\vec{AB} from the origin (0,0,0)(0,0,0) to point (1,0,2)(1,0,2): AB⃗=(1−0,0−0,2−0)=(1,0,2)\vec{AB} = (1 - 0, 0 - 0, 2 - 0) = (1, 0, 2)
Vector AC⃗\vec{AC} from the origin (0,0,0)(0,0,0) to point (−1,1,−2)(-1,1,-2): AC⃗=(−1−0,1−0,−2−0)=(−1,1,−2)\vec{AC} = (-1 - 0, 1 - 0, -2 - 0) = (-1, 1, -2)

STEP 4

Now, we calculate the cross product of vectors AB⃗\vec{AB} and AC⃗\vec{AC} to find the normal vector N⃗\vec{N} of the plane.
N⃗=AB⃗×AC⃗\vec{N} = \vec{AB} \times \vec{AC}

STEP 5

The cross product of two vectors (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) is given by:
∣ijkx1y1z1x2y2z2∣\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ \end{vmatrix}

STEP 6

We apply the cross product formula to vectors AB⃗\vec{AB} and AC⃗\vec{AC}:
N⃗=∣ijk102−11−2∣\vec{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2 \\ -1 & 1 & -2 \\ \end{vmatrix}

STEP 7

Calculate the determinant to find the components of the normal vector N⃗\vec{N}:
N⃗=(0⋅(−2)−2⋅1)i−(1⋅(−2)−2⋅(−1))j+(1⋅1−0⋅(−1))k\vec{N} = (0 \cdot (-2) - 2 \cdot 1) \mathbf{i} - (1 \cdot (-2) - 2 \cdot (-1)) \mathbf{j} + (1 \cdot 1 - 0 \cdot (-1)) \mathbf{k}

STEP 8

Simplify the components of the normal vector N⃗\vec{N}:
N⃗=(−2)i−(−2)j+(1)k\vec{N} = (-2) \mathbf{i} - (-2) \mathbf{j} + (1) \mathbf{k}
N⃗=(−2,2,1)\vec{N} = (-2, 2, 1)

STEP 9

The components of the normal vector N⃗\vec{N} give us the coefficients aa, bb, and cc for the equation of the plane:
a=−2,b=2,c=1a = -2, \quad b = 2, \quad c = 1

STEP 10

Substitute the values of aa, bb, and cc into the general equation of the plane:
−2x+2y+z=0-2x + 2y + z = 0
This is the equation of the plane that contains the given points.

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