ind an equation of the plane.

the plane through the point (6,2,9) and with normal vector2**i** +6**j** +5**k**

The equation of a plane can be written in the general form:

\[
A(x - x_1) + B(y - y_1) + C(z - z_1) = 0
\]

where \( (x_1, y_1, z_1) \) is a point on the plane, and \( A, B, C \) are the components of the normal vector to the plane.

Given the point \( (6, 2, 9) \) and the normal vector \( 2\mathbf{i} + 6\mathbf{j} + 5\mathbf{k} \), the components of the normal vector are \( A = 2 \), \( B = 6 \), and \( C = 5 \).

Substitute these values into the general equation:

\[
2(x - 6) + 6(y - 2) + 5(z - 9) = 0
\]

Now, simplify the equation:

\[
2(x - 6) + 6(y - 2) + 5(z - 9) = 0
\]
\[
2x - 12 + 6y - 12 + 5z - 45 = 0
\]
\[
2x + 6y + 5z - 69 = 0
\]

So, the equation of the plane is:

\[
2x + 6y + 5z = 69
\]

Would you like more details on how this was derived, or do you have any questions?

---

Here are five related questions to expand on this topic:
1. How do you determine the normal vector of a plane given three points?
2. What is the geometric interpretation of the normal vector in relation to the plane?
3. How can you determine if a point lies on the plane given its equation?
4. How can you find the angle between two planes given their normal vectors?
5. What is the equation of a plane in parametric form?

**Tip:** The normal vector defines the orientation of the plane in space and is perpendicular to the plane.

Find an equation of the plane. The plane through the point (6, 2, 9) and with normal vector 2i + 6j + 5k.

Learn how to find the equation of a plane passing through the point (6, 2, 9) with a normal vector of 2i + 6j + 5k. This problem demonstrates key concepts in vector calculus and geometry, ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation