222k views
0 votes
Find an equation of the plane which contains both the point p (1, 3, 2) and the line with symmetric equations x - 1 = -y/2 = z - 3/2?

User AndyC
by
7.6k points

1 Answer

4 votes

Final answer:

To find the equation of the plane that contains a point and a line, you need to identify the direction vector of the line and use it along with the given point to find the equation of the plane.

Step-by-step explanation:

To find the equation of the plane that contains both the point P(1, 3, 2) and the line with symmetric equations x - 1 = -y/2 = z - 3/2, we need to first identify the direction vector of the line. From the symmetric equations, we can see that the direction vector of the line is (1, -2, 1). We can use this direction vector along with the given point to find the equation of the plane.

The equation of the plane is given by the formula: A(x - x1) + B(y - y1) + C(z - z1) = 0, where (x1, y1, z1) is the given point, and (A, B, C) is the direction vector of the plane.

Substituting the values, we have (1 - x1) + (-2)(y - y1) + (1)(z - z1) = 0. Simplifying this equation will give us the equation of the plane.

User Bohdi
by
7.7k points