Let be the line of intersection of the planes given by the equationsLet be the line passing through the point and parallel to . Let denote the plane given by the equationSuppose that the line meets the plane at the point . Let be the foot of the perpendicular drawn from to the plane .If the length of the line segment is and the area of is , find the value of 4$$\ell$$^2 A^2.\
[THOUGHT] I'll:
- Find direction vector of L1 as cross product of normals of the two planes.
- Use that as direction for L2 through P, parametrize L2.
- Intersect L2 with plane Π to find Q.
- Compute foot of perpendicular R from P to Π using projection onto normal.
- Compute ℓ = |PQ| and A = area of triangle PQR = 0.5 * |PQ × PR|.
- Compute 4 * ℓ^2 * A^2 using Python for exact arithmetic. [/THOUGHT] [PYTHON] import sympy as sp
Define symbols
t = sp.symbols('t', real=True)
Given data
P = sp Point(2, -1, 3) n1 = sp.Matrix([2,3,1]) n2 = sp.Matrix([1,2,1])
Plane Pi: 2x + y - 2z = 6
nPi = sp.Matrix([2,1,-2]) dPi = 6
1. Direction vector of L1: cross of n1 and n2
v = n1.cross(n2) v
print("Direction vector v:", v)
2. L2: line through P parallel to v
x, y, z = sp.symbols('x y z', real=True) X = sp.Matrix([x,y,z])
Parametric L2: X = P + t*v
X_param = sp.Matrix(P) + t*v
3. Intersection with plane Pi: solve 2x + y - 2z = 6
eq_plane = nPi.dot(X_param) - dPi t_solution = sp.solve(sp.Eq(eq_plane,0), t)[0] t_solution_simplified = sp.simplify(t_solution) print("t for intersection:", t_solution_simplified)
Q = sp.Matrix(P) + t_solution_simplified*v Q_simplified = sp.simplify(Q) print("Point Q:", Q_simplified)
4. Foot of perpendicular R from P to plane Pi
Formula: R = P - ((n·P - d)/|n|^2) * n, for plane n·X = d
P_vec = sp.Matrix(P) lambda_factor = (nPi.dot(P_vec) - dPi) / (nPi.dot(nPi)) R = P_vec - lambda_factor * nPi R_simplified = sp.simplify(R) print("Point R:", R_simplified)
5. Compute ℓ = |PQ| and area A of triangle PQR
PQ = Q_simplified - P_vec PR = R_simplified - P_vec
ell = sp.sqrt(PQ.dot(PQ)) print("ell^2:", sp.simplify(PQ.dot(PQ)))
cross_PQ_PR = PQ.cross(PR) area = sp.Rational(1,2) * sp.sqrt(cross_PQ_PR.dot(cross_PQ_PR)) print("Area^2:", sp.simplify((cross_PQ_PR.dot(cross_PQ_PR))/4))
6. Compute 4 * ell^2 * A^2
ell2 = sp.simplify(PQ.dot(PQ)) A2 = sp.simplify((cross_PQ_PR.dot(cross_PQ_PR))/4) expr = sp.simplify(4 * ell2 * A2) print("4 * ell^2 * A^2 =", expr)
[THOUGHT]
The previous code had a syntax error because I wrote sp Point instead of sp.Point. I'll fix that and rerun the computation for the direction vector, intersection point Q, foot R, and then compute 4 * ell^2 * A^2.
[/THOUGHT]
[PYTHON]
import sympy as sp
Define symbols
t = sp.symbols('t', real=True)
Given data
P = sp.Point(2, -1, 3) n1 = sp.Matrix([2,3,1]) n2 = sp.Matrix([1,2,1])
Plane Pi: 2x + y - 2z = 6
nPi = sp.Matrix([2,1,-2]) dPi = 6
1. Direction vector of L1: cross of n1 and n2
v = n1.cross(n2) print("Direction vector v:", v)
2. L2: line through P parallel to v
P_vec = sp.Matrix(P) X_param = P_vec + t*v
3. Intersection with plane Pi: solve 2x + y - 2z = 6
eq_plane = nPi.dot(X_param) - dPi t_solution = sp.solve(sp.Eq(eq_plane,0), t)[0] t_solution_simplified = sp.simplify(t_solution) print("t for intersection:", t_solution_simplified)
Q = P_vec + t_solution_simplified*v Q_simplified = sp.simplify(Q) print("Point Q:", Q_simplified)
4. Foot of perpendicular R from P to plane Pi
Formula: R = P - ((n·P - d)/|n|^2) * n, for plane n·X = d
lambda_factor = (nPi.dot(P_vec) - dPi) / (nPi.dot(nPi)) R = P_vec - lambda_factor * nPi R_simplified = sp.simplify(R) print("Point R:", R_simplified)
5. Compute ℓ = |PQ| and area A of triangle PQR
PQ = Q_simplified - P_vec PR = R_simplified - P_vec
ell2 = sp.simplify(PQ.dot(PQ)) print("ell^2:", ell2)
cross_PQ_PR = PQ.cross(PR) A2 = sp.simplify((cross_PQ_PR.dot(cross_PQ_PR))/4) print("Area^2:", A2)
6. Compute 4 * ell^2 * A^2
expr = sp.simplify(4 * ell2 * A2) print("4 * ell^2 * A^2 =", expr)
[THOUGHT] Using the valid Python output, the value of (4\ell^2 A^2) is 511758. [ \boxed{511758} ]