Equation of a Plane

To find the equation of a plane, we need a point on the plane and the normal vector to the plane. In this case, we are given that the normal to the plane has a length of 5 and makes angles of 120°, 45°, and 120° with the x, y, and z axes respectively.

Normal Vector

The normal vector to the plane can be found by using the direction cosines of the angles. The direction cosines of an angle are the cosines of the angles it makes with the x, y, and z axes. Let's denote the direction cosines as l, m, and n for the x, y, and z axes respectively.

Given that the angles with the axes are 120°, 45°, and 120°, we can determine the direction cosines:

l = cos(120°) = -0.5
m = cos(45°) = 0.7071
n = cos(120°) = -0.5

Now, we can find the normal vector by multiplying the direction cosines by the length of the normal vector:

N = 5 * (-0.5)i + 5 * 0.7071j + 5 * (-0.5)k

Simplifying, we get:

N = -2.5i + 3.5355j - 2.5k

Equation of the Plane

Now that we have the normal vector, we need a point on the plane. Since the normal vector is from the origin, the plane passes through the origin (0, 0, 0).

The equation of a plane in vector form is given by:

N · (r - r0) = 0

Where N is the normal vector, r is a general position vector on the plane, and r0 is a known point on the plane (in this case, the origin).

Substituting the values, we get:

(-2.5i + 3.5355j - 2.5k) · (xi + yj + zk - 0) = 0

Simplifying, we get:

-2.5x + 3.5355y - 2.5z = 0

Thus, the equation of the plane is:

-2.5x + 3.5355y - 2.5z = 0