Discrete Math: Finding Angle Between Plane & XZ Axis

AI Thread Summary
To find the angle between a plane and the xz-axis, one must consider the normals of the planes. The angle between two planes is determined by the angle between their normal vectors. For a plane represented by Ax + By + Cz = D, the normal vector is Ai + Bj + Ck, while the xz-plane has a normal vector of j. The cosine of the angle θ between the planes can be calculated using the formula cos(θ) = B / √(A² + B² + C²). This approach provides a clear method for determining the angle between the specified plane and the xz-axis.
eme_girl
Messages
5
Reaction score
0
How do you find the angle between the co-ordinate axis (i.e. the xz plane) and another plane in general?
 
Physics news on Phys.org
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 cos(\theta)= \frac{B}{\sqrt{A^2+B^2+C^2}}.
 
Last edited by a moderator:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top