Find the equations of the plane the normal to which from the origin is...
Equation of a Plane with Given Normal and Angles
To find the equation of the plane with the given normal and angles, we need to follow these steps:
Step 1: Find the components of the normal vector
Let n be the normal vector of the plane. We know that the length of n is 5, so:
$$|\textbf{n}|=5$$
Since the angle between n and the x-axis is 120 degrees, we can find the x-component of n using:
$$n_x=|\textbf{n}|\cos(120^\circ)=-\frac{5}{2}$$
Similarly, we can find the y- and z-components of n:
$$n_y=|\textbf{n}|\cos(45^\circ)=\frac{5}{\sqrt{2}}$$
$$n_z=|\textbf{n}|\cos(120^\circ)=-\frac{5}{2}$$
Therefore, the normal vector of the plane is:
$$\textbf{n}=\begin{pmatrix}-\frac{5}{2}\\\frac{5}{\sqrt{2}}\\-\frac{5}{2}\end{pmatrix}$$
Step 2: Find the equation of the plane
Let r be any point on the plane, and let r0 be the position vector of a point on the plane. The equation of the plane can be written as:
$$\textbf{n}\cdot(\textbf{r}-\textbf{r}_0)=0$$
Expanding the dot product, we get:
$$-\frac{5}{2}(x-x_0)+\frac{5}{\sqrt{2}}(y-y_0)-\frac{5}{2}(z-z_0)=0$$
Dividing by 5, we get:
$$-\frac{1}{2}(x-x_0)+\frac{1}{\sqrt{2}}(y-y_0)-\frac{1}{2}(z-z_0)=0$$
This is the equation of the plane with the given normal and angles.