Discrete Math: Finding Angle Between Plane & XZ Axis

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SUMMARY

The angle between the xz-plane and another plane can be determined using the normals of the planes. For a plane defined by the equation Ax + By + Cz = D, the normal vector is represented as Ai + Bj + Ck. The angle θ between the xz-plane (with normal vector j) and the given plane is calculated using the formula cos(θ) = B / √(A² + B² + C²), where B is the coefficient of y in the plane's equation.

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  • Understanding of vector mathematics
  • Familiarity with the concept of normal vectors
  • Knowledge of dot product calculations
  • Basic grasp of trigonometric functions
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  • Study vector calculus and its applications in geometry
  • Learn about the properties of normal vectors in three-dimensional space
  • Explore advanced trigonometric identities and their applications
  • Investigate the geometric interpretation of dot products
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Students and professionals in mathematics, physics, and engineering who need to calculate angles between planes in three-dimensional space.

eme_girl
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How do you find the angle between the co-ordinate axis (i.e. the xz plane) and another plane in general?
 
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eme_girl said:
How do you find the angle between the co-ordinate axis (i.e. the xz plane) and another plane in general?

Angle between two planes is the angle between the normals of the planes.
 
Expanding on what learningphysics said, if one plane is given by Ax+ By+ Cz= D and the other by ax+ by+cz= d, then the normal vectors are Ai+ Bj+ Ck and ai+ bj+ ck respectively. u.v= |u||v|cos(θ) so θ, the angle between the two vectors and the angle between the planes, is given by cos(θ)= u.v/(|u||v|).


In particular, the xz-plane has normal vector j. If the other plane is given by Ax+By+Cz= D, its normal vector is Ai+Bj+Ck. The dot product of those is simply B so the angle between the planes is given by [itex]cos(\theta)= \frac{B}{\sqrt{A^2+B^2+C^2}}[/itex].
 
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