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] Compute direction vector of L1 as cross product of plane normals. Parametrize L2 through P with that direction, intersect with plane Π to get Q. Find R as orthogonal projection of P onto Π. Then ℓ=|PQ| and area A = 1/2 |(Q-P)×(R-P)|. Finally compute 4ℓ^2A^2 (which simplifies to |PQ|^2 * |(Q-P)×(R-P)|^2). [/THOUGHT] [PYTHON] import sympy as sp
Define normals of the planes for L1
n1 = sp.Matrix([2,3,1]) n2 = sp.Matrix([1,2,1])
Direction of intersection line L1
d = n1.cross(n2) d print("d =", d)
Point P
P = sp.Matrix([2,-1,3])
Plane Pi: 2x + y - 2z = 6
nPi = sp.Matrix([2,1,-2])
Parametric line L2: P + t*d
t = sp.Symbol('t', real=True) X = P + t*d
Solve for intersection with plane Pi
eq = sp.Eq(2X[0] + X[1] - 2X[2], 6) t_sol = sp.solve(eq, t)[0] Q = sp.simplify(X.subs(t, t_sol))
print("t_sol =", t_sol) print("Q =", Q)
Foot of perpendicular from P to plane Pi
Plane: nPi . x = 6, so projection: R = P - ((n.P - 6)/||n||^2) * n
n_norm_sq = (nPi.dot(nPi)) lam = (nPi.dot(P) - 6) / n_norm_sq R = sp.simplify(P - lam*nPi)
print("R =", R)
Compute length PQ
PQ = Q - P ell_sq = sp.simplify(PQ.dot(PQ)) ell = sp.sqrt(ell_sq)
Area of triangle PQR
PR = R - P cross = PQ.cross(PR) cross_sq = sp.simplify(cross.dot(cross)) A_sq = sp.simplify(cross_sq/4)
expr = sp.simplify(4ell_sqA_sq) print("ell_sq =", ell_sq) print("cross_sq =", cross_sq) print("A_sq =", A_sq) print("4ell^2A^2 =", expr) print("as integer?", sp.nsimplify(expr)) [/PYTHON]
[THOUGHT] Using the computed values, (4\ell^2A^2 = 511758). [\boxed{511758}]