ANSWERS: 1
  • Let Θ1 = the angle of the vector on the x-y plane. Let Θ2 = the angle of the vector on the y-z plane. Let m = the magnitude of the vector. Let x1 = the projection distance of the vector on the x plane. Let y1 = the projection distance of the vector on the y plane. Let z1 = the projection distance of the vector on the z plane. x1 = m{cos(Θ1)} y1 = m{sin(Θ1)} z1 = m{sin(Θ2)} This means that the vector's direction is: {cos(Θ1)i, sin(Θ1)j, sin(Θ2)k} To turn this into a unit vector you must divide by: sqrt{cos^2(Θ1) + sin^2(Θ1) + sin^2(Θ2)} so a unit vector in the direction of the vector is: {cos(Θ1)i, sin(Θ1)j, sin(Θ2)k}/sqrt{cos^2(Θ1) + sin^2(Θ1) + sin^2(Θ2)} now multiply by the magnitude of the vector: m{cos(Θ1)i, sin(Θ1)j, sin(Θ2)k}/sqrt{cos^2(Θ1) + sin^2(Θ1) + sin^2(Θ2)} This assumes that the vector's tail is at the orgin.

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