Find the distance between the point Q = (3,2,-3) and x - 2y + z = 5

To find the distance between a point \( Q = (3, 2, -3) \) and a plane given by the equation \( x - 2y + z = 5 \), we use the formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \):

\[
D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]

### Step 1: Rewrite the plane equation in standard form
The given equation is:
\[
x - 2y + z = 5
\]
which can be rewritten as:
\[
x - 2y + z - 5 = 0
\]
Here, \( A = 1 \), \( B = -2 \), \( C = 1 \), and \( D = -5 \).

### Step 2: Apply the formula
Substitute the coordinates of point \( Q = (3, 2, -3) \) into the formula:

\[
D = \frac{|(1)(3) + (-2)(2) + (1)(-3) - 5|}{\sqrt{1^2 + (-2)^2 + 1^2}}
\]

### Step 3: Simplify the numerator
\[
D = \frac{|3 - 4 - 3 - 5|}{\sqrt{1 + 4 + 1}} = \frac{|-9|}{\sqrt{6}} = \frac{9}{\sqrt{6}}
\]

### Step 4: Simplify the result
To simplify further, multiply the numerator and denominator by \( \sqrt{6} \):

\[
D = \frac{9\sqrt{6}}{6} = \frac{3\sqrt{6}}{2}
\]

Thus, the distance between the point \( Q \) and the plane is:
\[
D = \frac{3\sqrt{6}}{2}
\]

Would you like more details on any step or have any questions?

Here are 5 related questions to extend this topic:
1. What is the distance from a point to a line in 3D space?
2. How do you find the equation of a plane given three points?
3. What is the geometric interpretation of the dot product in this context?
4. How do you calculate the projection of a vector onto a plane?
5. How do you find the intersection of a line and a plane?

**Tip:** When solving distance problems involving planes, always ensure the equation is in standard form before applying the formula.

Learn how to find the distance from the point Q = (3, 2, -3) to the plane x - 2y + z = 5 using the formula for distance in 3D geometry. This detailed explanation includes step-by-step calculations and simplifies the result for easier understanding. Suitable for students in Grades 10-12.

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