. A triangle is formed by the points P = (1, 1, 3), Q = (4, −1, 3), and R = (7, 1, −2).
(a) Sketch the triangle.
(b) Find the vector and scalar equations of the plane containing this triangle

Let's solve the problem step by step. Here's how we approach each part:

### (a) Sketch the triangle

Since this is a 3D triangle, a direct sketch would involve plotting points in 3D space. The three points of the triangle are:

- \( P(1, 1, 3) \)
- \( Q(4, -1, 3) \)
- \( R(7, 1, -2) \)

Although I cannot generate a 3D sketch here, I can describe the process:

1. **Axes**: Draw the three coordinate axes \( x \), \( y \), and \( z \).
2. **Plot points**:
   - \( P = (1, 1, 3) \): Move 1 unit along the x-axis, 1 unit along the y-axis, and 3 units along the z-axis.
   - \( Q = (4, -1, 3) \): Move 4 units along the x-axis, -1 unit (down) along the y-axis, and 3 units along the z-axis.
   - \( R = (7, 1, -2) \): Move 7 units along the x-axis, 1 unit along the y-axis, and -2 units (down) along the z-axis.
3. **Connect the points** to form the triangle.

If you'd like a graphical representation, I can assist you with that as well.

### (b) Find the vector and scalar equations of the plane

#### 1. **Vector equation of the plane**

To find the vector equation, we need a point on the plane and two direction vectors that lie on the plane.

- Choose \( P(1, 1, 3) \) as the reference point.
- Find two direction vectors using the points \( Q \) and \( R \):

  \[
  \overrightarrow{PQ} = Q - P = (4 - 1, -1 - 1, 3 - 3) = (3, -2, 0)
  \]
  \[
  \overrightarrow{PR} = R - P = (7 - 1, 1 - 1, -2 - 3) = (6, 0, -5)
  \]

The vector equation of a plane is given by:
\[
\mathbf{r} = \mathbf{P} + s \overrightarrow{PQ} + t \overrightarrow{PR}
\]
Substituting the values of \( P \), \( \overrightarrow{PQ} \), and \( \overrightarrow{PR} \):

\[
\mathbf{r} = (1, 1, 3) + s(3, -2, 0) + t(6, 0, -5)
\]

Thus, the vector equation is:
\[
\mathbf{r} = (1 + 3s + 6t, 1 - 2s, 3 - 5t)
\]

#### 2. **Scalar equation of the plane**

The scalar equation of a plane is of the form:
\[
Ax + By + Cz = D
\]
To find \( A \), \( B \), and \( C \), we need the normal vector \( \mathbf{n} \) to the plane. The normal vector is obtained by taking the cross product of the direction vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \):

\[
\mathbf{n} = \overrightarrow{PQ} \times \overrightarrow{PR}
\]
\[
\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 0 \\ 6 & 0 & -5 \end{vmatrix}
\]

Let's compute this:

\[
\mathbf{n} = \mathbf{i}((-2)(-5) - (0)(0)) - \mathbf{j}((3)(-5) - (0)(6)) + \mathbf{k}((3)(0) - (-2)(6))
\]
\[
\mathbf{n} = \mathbf{i}(10) - \mathbf{j}(-15) + \mathbf{k}(12)
\]
\[
\mathbf{n} = (10, 15, 12)
\]

Thus, the normal vector to the plane is \( \mathbf{n} = (10, 15, 12) \).

Now, use the point \( P(1, 1, 3) \) to find \( D \) in the scalar equation. The scalar equation is given by:
\[
10(x - 1) + 15(y - 1) + 12(z - 3) = 0
\]

Expanding this:
\[
10x - 10 + 15y - 15 + 12z - 36 = 0
\]
\[
10x + 15y + 12z = 61
\]

Thus, the scalar equation of the plane is:
\[
10x + 15y + 12z = 61
\]

### Summary of Results:
1. **Vector equation of the plane**:
   \[
   \mathbf{r} = (1 + 3s + 6t, 1 - 2s, 3 - 5t)
   \]
2. **Scalar equation of the plane**:
   \[
   10x + 15y + 12z = 61
   \]

Would you like further details or clarifications on any part of this? Here are some related questions:

1. How is the cross product used to find the normal vector to a plane?
2. How can you interpret the scalar equation of a plane geometrically?
3. What is the difference between a vector and scalar equation in 3D geometry?
4. How do you check if a point lies on a given plane?
5. What other forms of the equation of a plane can we use in 3D space?

**Tip:** To verify your results, substitute the coordinates of \( P \), \( Q \), and \( R \) into the scalar equation to ensure that they all satisfy it.

A triangle is formed by the points P = (1, 1, 3), Q = (4, −1, 3), and R = (7, 1, −2). (a) Sketch the triangle. (b) Find the vector and scalar equations of the plane containing this triangle.

Learn how to sketch a triangle in 3D and find both the vector and scalar equations of the plane containing points P(1,1,3), Q(4,-1,3), and R(7,1,-2). This problem covers key concepts in 3D geometry, including vector equations and cross products, suitable for high school and college-level students.

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