To solve the problem in the image, we need to determine the equation of the plane that passes through the given points $P(1, -1, 3)$, $Q(2, -1, 2)$, and $R(3, 0, 3)$.

Step 1: Find the vectors

We need two vectors that lie on the plane. These can be obtained from the given points:

• Vector $\mathbf{PQ} = Q - P = (2 - 1, -1 - (-1), 2 - 3) = (1, 0, -1)$
• Vector $\mathbf{PR} = R - P = (3 - 1, 0 - (-1), 3 - 3) = (2, 1, 0)$

Step 2: Find the normal vector

The normal vector to the plane can be found by taking the cross product of $\mathbf{PQ}$ and $\mathbf{PR}$.

$\mathbf{n} = \mathbf{PQ} \times \mathbf{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{vmatrix}$

$\mathbf{n} = \mathbf{i} \left( 0(0) - (-1)(1) \right) - \mathbf{j} \left( 1(0) - (-1)(2) \right) + \mathbf{k} \left( 1(1) - 0(2) \right)$

$\mathbf{n} = \mathbf{i}(1) - \mathbf{j}(2) + \mathbf{k}(1)$

So the normal vector is $\mathbf{n} = (1, -2, 1)$.

Step 3: Use the normal vector to find the plane equation

The general form of the plane equation is: $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$ where $(x_1, y_1, z_1)$ is a point on the plane (e.g., point $P(1, -1, 3)$) and $\mathbf{n} = (a, b, c) = (1, -2, 1)$.

Substitute the values: $1(x - 1) - 2(y + 1) + 1(z - 3) = 0$ Simplifying: $x - 1 - 2y - 2 + z - 3 = 0$ $x - 2y + z - 6 = 0$

Step 4: Conclusion

The equation of the plane is $x - 2y + z = 6$, which corresponds to the second option in the provided choices: $x - 2y + z - 6 = 0$

Would you like further details or have any questions?

Related Questions:

1. How do you calculate the cross product of two vectors?
2. What is the geometric meaning of the normal vector to a plane?
3. How do you find the equation of a plane using three points in space?
4. How do you verify if a point lies on a plane?
5. What is the significance of the coefficients in the equation of a plane?

Tip:

When solving plane equations, always check the normal vector, as it determines the orientation of the plane in space.