Q2. Find the equation of the plane containing the point (2,1,1) and the line r(t)=⟨4+t, 2-t, 2+t⟩. there is half arrow on top of r

To find the equation of the plane that contains a point and a line, we can use the following steps:

1. Identify a point on the plane.
2. Find two vectors that lie on the plane.
3. Use the normal vector of the plane, which is perpendicular to the two vectors found in step 2, to write the equation of the plane.

Given:
- A point \( P = (2, 1, 1) \)
- A line \( \mathbf{r}(t) = \langle 4+t, 2-t, 2+t \rangle \)

### Step 1: Identify a Point on the Plane
The point \( P = (2, 1, 1) \) is already given as lying on the plane.

### Step 2: Find Two Vectors on the Plane
First, we identify a point on the line by substituting \( t = 0 \):
\[ \mathbf{r}(0) = \langle 4, 2, 2 \rangle \]
Let's call this point \( Q = (4, 2, 2) \).

Next, identify another point on the line by substituting \( t = 1 \):
\[ \mathbf{r}(1) = \langle 5, 1, 3 \rangle \]
Let's call this point \( R = (5, 1, 3) \).

Now, we can find two vectors that lie on the plane:
- Vector \(\mathbf{PQ}\) from \( P \) to \( Q \):
\[ \mathbf{PQ} = Q - P = (4 - 2, 2 - 1, 2 - 1) = (2, 1, 1) \]

- Vector \(\mathbf{PR}\) from \( P \) to \( R \):
\[ \mathbf{PR} = R - P = (5 - 2, 1 - 1, 3 - 1) = (3, 0, 2) \]

### Step 3: Find the Normal Vector to the Plane
To find the normal vector \(\mathbf{n}\) to the plane, we take the cross product of vectors \(\mathbf{PQ}\) and \(\mathbf{PR}\):
\[ \mathbf{PQ} \times \mathbf{PR} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 1 \\
3 & 0 & 2
\end{vmatrix} \]

Calculating the cross product:
\[
\mathbf{n} = \mathbf{i} (1 \cdot 2 - 1 \cdot 0) - \mathbf{j} (2 \cdot 2 - 1 \cdot 3) + \mathbf{k} (2 \cdot 0 - 1 \cdot 3)
\]
\[
\mathbf{n} = \mathbf{i} (2) - \mathbf{j} (4 - 3) + \mathbf{k} (-3)
\]
\[
\mathbf{n} = \langle 2, -1, -3 \rangle
\]

### Step 4: Write the Equation of the Plane
The equation of the plane is given by:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

Using the normal vector \(\mathbf{n} = \langle 2, -1, -3 \rangle\) and the point \( P = (2, 1, 1) \):
\[ 2(x - 2) - 1(y - 1) - 3(z - 1) = 0 \]
\[ 2x - 4 - y + 1 - 3z + 3 = 0 \]
\[ 2x - y - 3z = 0 \]

So, the equation of the plane is:
\[ 2x - y - 3z = 0 \]

Would you like more details or have any questions?

### Related Questions:
1. How do you find the distance between a point and a plane?
2. What is the equation of a plane given three points?
3. How can you determine if two planes are parallel?
4. How do you find the intersection line of two planes?
5. What is the method to find the angle between two planes?
6. How can you verify if a given point lies on a plane?
7. What is the significance of the normal vector in the equation of a plane?
8. How do you convert the parametric equation of a line into the symmetric form?

### Tip:
Remember that the normal vector of a plane provides valuable information about the plane's orientation in space, and it is essential for defining the plane's equation.

Learn how to find the equation of a plane that contains a given point and a parametric line. This detailed step-by-step solution involves vector calculations and the cross product method. Suitable for undergraduate level mathematics.

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