To find the equation of a plane that contains a given line and is perpendicular to a specified plane, we need to follow a structured series of steps. This article will guide you through the process with a detailed example and explain the underlying mathematical principles.
Step-by-Step Guide
1. Determine the Direction Vector of the Line
The given line is in parametric form:[ x -2t, y 2 - 4t, z 42t ]
We can extract the direction vector of the line from the coefficients of (t):
( mathbf{d} begin{pmatrix} 1 -4 2 end{pmatrix} )
2. Find the Normal Vector of the Given Plane
The equation of the given plane is:[ x - y - z 9 ]
Converting it to the general form ( Ax By Cz D ), we get:
( A 1, B -1, C -1, D 9 )
The normal vector ( mathbf{n} ) of the plane is:
( mathbf{n} begin{pmatrix} 1 -1 -1 end{pmatrix} )
3. Determine the Normal Vector of the Desired Plane
The normal vector of the desired plane must be perpendicular to the normal vector of the given plane and the direction vector of the line. Therefore, the normal vector ( mathbf{N} ) can be found by taking the cross product of ( mathbf{d} ) and ( mathbf{n} ):[ mathbf{N} mathbf{d} times mathbf{n} ]
Calculating the cross product:
( mathbf{N} begin{vmatrix} hat{i} hat{j} hat{k} 1 -4 2 1 -1 -1 end{vmatrix} hat{i} begin{vmatrix} -4 2 -1 -1 end{vmatrix} - hat{j} begin{vmatrix} 1 2 1 -1 end{vmatrix} hat{k} begin{vmatrix} 1 -4 1 -1 end{vmatrix} )
( hat{i} (-4 2) - hat{j} (-1 - 2) hat{k} (-1 4) )
( -2 hat{i} 3 hat{j} 3 hat{k} )
( begin{pmatrix} -2 1 3 end{pmatrix} )
Thus, the normal vector of the plane we are seeking is:
( mathbf{N} begin{pmatrix} -2 1 3 end{pmatrix} )
4. Write the Equation of the Plane
The general form of the plane equation is:( -2x - (2)x_0 1y - (2)y_0 3z - (2)z_0 0 )
We can choose a point on the given line by setting ( t 0 ):
[ x -2, y 2, z 4 ]
Substituting these values into the plane equation:
( -2x - 2 1y - 2 3z - 4 0 )
Simplifying the equation:
( -2x y 3z - 8 0 )
Rearranging the terms gives us the final equation of the plane:
( 2x - y - 3z -18 )
Conclusion
By following these steps, we have determined the equation of a plane that is perpendicular to the given plane and contains the specified line. The final equation is:( boxed{2x - y - 3z -18} )