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If E1 = a1x + b1y + c1z + d = 0, E2 = a2x + b2y + c2z + d = 0, are two intersecting planes which intersect in a line L, then the equation of any plane passing through line L is of the form λ1E1 + λ2E2 = 0 where |λ1| + |λ2| ≠ 0. |
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Answer» Suppose E ≡ px + qy + rz + t = 0 is a plane containing the line L. Let vector n1 = (a1, b1, c1) and vector n2 (a2, b2, c2) be the normals to E1 = 0 and E2 = 0, respectively. Hence, vector (n1 x n2) is parallel to the line L so that (b1c2 - b2c1 - c2a1, a1b2 - a2b1) are the DRs of the line L. Since one of the DRs is not zero, we consider that a1b2 - a2b1 ≠ 0. Therefore, the equations a1x + a2y = p and b1x + b2y = q are uniquely solvable for a non-zero solution and let x = λ1 and y1 = λ2 be the solution. Therefore, a1λ1 + a2λ2 = p and b1λ1 + b2λ2 = q. Let k = r - λ1c1 - λ2c2 and l = t - λ1d1 - λ2d2 so that E ≡ λ1(a1x + b1y + c1z + d1) + λ2(a2x + b2y + c2z + d2) + kz + l = 0 ...(1) Suppose P (x1, y1, z1) and Q(x2, y2, z2), where z1 ≠ z2, are points on line L. Therefore, a1x1 + b1y1 + c1z1 + d1 = 0 and a2x1 + b2y1 + c2z1 + d2 = 0 so that point P(x1, y1, z1) satisfies the equation E = 0 and similarly point Q(x2, y2, z2) satisfies E = 0 [from Eq. (1)]. Therefore, kz1 + l = 0, kz2 + l = 0 and z1 ≠ z2 k = 0 and then l = 0. Hence, the equation of the plane E = 0 is of the for λ1E1 + λ2E2 = 0. |
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