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The plane ax+by=0 is rotated about its line of intersection with the plane z=0 through an angle $\alpha$.Prove that the equation of the plane in its new position is $ax+by\pm \bigg(\sqrt {a^2+b^2}\tan \alpha\bigg)z=0$.

1 Answer

P1: ax+by=0 P2: z=0 Angle between p1 and p2 Cos(¢)= 0 ¢= π/2 P3/p4: ax+by+kz=0 When a plane is rotated thru an angle α Angle b/w p1 and p3/p4 Cosα= a^2+b^2/√(a^2+ b^2) [√a^2+b^2+k^2] Angle b/w p2 and p3/p4 Cos(π/2-α)= k/ [√a^2+b^2+k^2] Tanα= k/ √(a^2+ b^2) K= √(a^2+ b^2) tanα Plane =ax +by + √(a^2+ b^2)tanα
answered Feb 4 by sanjamdeep30

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