Suppose the plane x+ay+bz=cx+ay+bz=c contains the point (1,2,3)(1,2,3) and the line ℓ:r(t)=(3,9,8)+t(1,1,1)ℓ:r(t)=(3,9,8)+t(1,1,1), t∈Rt∈R. Find a,b and c.

To find the values of \( a \), \( b \), and \( c \) for the plane equation \( x + ay + bz = c \) that contains the point \( (1, 2, 3) \) and the line \( \ell: \mathbf{r}(t) = (3, 9, 8) + t(1, 1, 1) \), we can proceed as follows:

### Step 1: Plane Contains the Point (1, 2, 3)

Since the plane contains the point \( (1, 2, 3) \), substituting these coordinates into the plane equation gives:

\[
1 + 2a + 3b = c.
\]

### Step 2: Plane Contains the Line

The line \( \ell: \mathbf{r}(t) = (3, 9, 8) + t(1, 1, 1) \) has a direction vector \( \mathbf{d} = (1, 1, 1) \).

Since the plane contains this line, the direction vector of the line must be perpendicular to the normal vector of the plane. The normal vector of the plane \( x + ay + bz = c \) is \( \mathbf{n} = (1, a, b) \).

For the plane to be perpendicular to the direction vector of the line, their dot product must be zero:

\[
\mathbf{n} \cdot \mathbf{d} = (1, a, b) \cdot (1, 1, 1) = 0.
\]

Calculating the dot product:

\[
1 \cdot 1 + a \cdot 1 + b \cdot 1 = 0.
\]

This simplifies to:

\[
1 + a + b = 0.
\]

### Step 3: Solve the System of Equations

We now have two equations:

1. \( 1 + 2a + 3b = c \).
2. \( 1 + a + b = 0 \).

We can use these equations to solve for \( a \), \( b \), and \( c \).

**From the second equation:**

\[
a + b = -1 \implies a = -1 - b.
\]

**Substitute \( a = -1 - b \) into the first equation:**

\[
1 + 2(-1 - b) + 3b = c.
\]

Simplify:

\[
1 - 2 - 2b + 3b = c,
\]

\[
-1 + b = c.
\]

So, we have:

\[
c = b - 1.
\]

Now, we have \( a = -1 - b \), \( c = b - 1 \), and the relation between \( a \) and \( b \).

### Step 4: Choosing a Value for \( b \)

Since \( b \) can be any real number, we can choose a specific value for simplicity. Let’s choose \( b = 0 \):

- \( b = 0 \)
- \( a = -1 - 0 = -1 \)
- \( c = 0 - 1 = -1 \)

So, the values are:

- \( a = -1 \)
- \( b = 0 \)
- \( c = -1 \)

### Conclusion

Thus, the coefficients of the plane are \( a = -1 \), \( b = 0 \), and \( c = -1 \). The equation of the plane is:

\[
x - y = -1.
\]

### Would you like more details or have any questions?

Here are 5 additional questions related to this topic:

1. How would the solution change if the point on the plane was different?
2. What if the plane also contained a different line, how would you find \( a \), \( b \), and \( c \)?
3. How can you determine if a point is on a given line or plane?
4. What methods can be used to find the intersection line of two planes?
5. How does the normal vector affect the orientation of the plane?

**Tip:** Verifying the solution by plugging back into the equations helps ensure accuracy.

Learn how to find the values of a, b, and c for the plane equation x + ay + bz = c that contains the point (1, 2, 3) and the line ℓ: r(t) = (3, 9, 8) + t(1, 1, 1). This problem involves vector mathematics, dot products, and plane equations, suitable for advanced high school students.

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