Solve the following problem by showing the minimum number of steps.

 Given three points $A(1, -1, 2)$, $B(1, 0, 1)$, and $C(-1, 0, 5)$.
\begin{ea}
\item Find the equation of the plane in 3-space which passes through $A$, $B$, and $C$.
\item Does the point $D(1, 2, 9)$ belong to the plane? If not find the distance from $D$ to the plane.
\end{ea}

Let's solve the problem with clear, concise steps.

### Part (a): Finding the Equation of the Plane

1. **Find two vectors on the plane:**
   \[
   \overrightarrow{AB} = B - A = (1 - 1, 0 + 1, 1 - 2) = (0, 1, -1)
   \]
   \[
   \overrightarrow{AC} = C - A = (-1 - 1, 0 + 1, 5 - 2) = (-2, 1, 3)
   \]

2. **Find the normal vector to the plane by taking the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:**
   \[
   \overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   0 & 1 & -1 \\
   -2 & 1 & 3
   \end{vmatrix}
   \]
   \[
   \overrightarrow{n} = \mathbf{i}(1 \cdot 3 - (-1) \cdot 1) - \mathbf{j}(0 \cdot 3 - (-1) \cdot (-2)) + \mathbf{k}(0 \cdot 1 - 1 \cdot (-2))
   \]
   \[
   \overrightarrow{n} = \mathbf{i}(3 + 1) - \mathbf{j}(0 - 2) + \mathbf{k}(0 + 2)
   \]
   \[
   \overrightarrow{n} = 4\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}
   \]
   Thus, the normal vector $\overrightarrow{n}$ is $(4, -2, 2)$.

3. **Form the equation of the plane using the point $A(1, -1, 2)$ and the normal vector:**
   \[
   4(x - 1) - 2(y + 1) + 2(z - 2) = 0
   \]
   Simplify the equation:
   \[
   4x - 4 - 2y - 2 + 2z - 4 = 0
   \]
   \[
   4x - 2y + 2z - 10 = 0
   \]

Thus, the equation of the plane is:
\[ 4x - 2y + 2z = 10 \]

### Part (b): Checking Point $D(1, 2, 9)$ and Finding the Distance if it Doesn't Lie on the Plane

1. **Check if $D(1, 2, 9)$ lies on the plane:**
   Substitute $D(1, 2, 9)$ into the plane equation $4x - 2y + 2z = 10$:
   \[
   4(1) - 2(2) + 2(9) = 4 - 4 + 18 = 18 \neq 10
   \]
   Since 18 does not equal 10, point $D$ does not lie on the plane.

2. **Find the distance from $D$ to the plane:**
   The distance from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is given by:
   \[
   \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
   \]
   For the plane $4x - 2y + 2z - 10 = 0$, we have $A = 4$, $B = -2$, $C = 2$, and $D = -10$. The point $D$ is $(1, 2, 9)$.
   \[
   \text{Distance} = \frac{|4(1) - 2(2) + 2(9) - 10|}{\sqrt{4^2 + (-2)^2 + 2^2}}
   \]
   \[
   \text{Distance} = \frac{|4 - 4 + 18 - 10|}{\sqrt{16 + 4 + 4}}
   \]
   \[
   \text{Distance} = \frac{|8|}{\sqrt{24}} = \frac{8}{2\sqrt{6}} = \frac{4}{\sqrt{6}} = \frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}
   \]

Thus, the distance from point $D(1, 2, 9)$ to the plane is $\frac{2\sqrt{6}}{3}$.

Would you like any additional details or explanations? Here are some questions you might be interested in:

1. How do you compute a cross product in vector algebra?
2. What are the steps to find the normal vector to a plane?
3. Can you explain the point-normal form of a plane equation?
4. How do you determine if a point lies on a given plane?
5. What is the formula to calculate the distance from a point to a plane?

Tip: When working with planes and vectors, visualizing the geometric relationships can be extremely helpful in understanding the concepts.

Learn how to find the equation of a plane passing through three points in 3-space: A(1, -1, 2), B(1, 0, 1), and C(-1, 0, 5). Also, check if point D(1, 2, 9) belongs to the plane and find the distance from D to the plane. This problem involves vector algebra, cross product, and plane equations. Suitable for students in Grades 11-12.

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