Respuesta :
Answer:
-3*x+6*y+3*z+9=0 (scalar equation of the plane)
Step-by-step explanation:
for P=(-4, -3, -1) and
L (x,y,z) = (2-3t , -7+6t ,-4+3t) = (2,7,-4)+(-3,6,3)*t = P₀ + v*t ( parametric equation of a line)
then v= (-3,6,3) is the vector parallel to the line and (2,-7,-4) is the point of the line at t=0
the equation for the plane could be solved in many ways
A) finding the point of the Line L₁, such that the vector L₁P is perpendicular to v , then the normal vector for the plane will be the vectorial product of L₁P and v
B) finding a point outside the plane P₂ such that P₂P is perpendicular to v
C) building a system of equations for the points that the plane contains ( t can be chosen in order to simplify the equations)
i will choose B)
P₂P= (x,y,z) -(-4, -3, -1) = x+4,y+3,z+1
then
P₂P*v =0
(x+4,y+3,z+1 )*(-3,6,3) = 0
(-3*x-12) + (6*y+ 18) + (3*z+3) = 0
-3*x+6*y+3*z+9=0 (scalar equation of the plane)
