How to determine if any point is in the acute or the obtuse angle between 2 planes

Question

Show that the point $(3, 2, -1)$ lies inside the acute angle formed by the planes $5x-y+z+3=0$ and $4x-3y+2z+5=0$.

I have tried this by calculating the angles between the plane, passing through the line of intersection of these two planes, and these planes. But the angles can't be calculated without calculator. So I need to get the solution in an easier way without calculator.

I also tried to get a method similar as the case of origin. To get if a origin is in the acute or the obtuse angle b/w 2 planes we check the sign of $a_1a_2+b_1b_2+c_1c_2$ if equations of the planes are $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2$ provided both $d_1$ and $d_2$ are positive.

I want to get this type of solution for this case also.

Answer 1

General plane passing through line of intersection of two planes $S_1=0$ and $S_2=0$ is $S_1+k*S_2=0$

In your case, the general plane is $5x-y+z+3+k(4x-3y+2z+5)=0$.

Since it passes through the given point $(3,2,-1)$, value of $k$ can be found.

Now add the two angles between new plane and two original planes to get the final answer.

Answer 2

Here is a systematic method to solve all such questions. In general we have two planes $P,Q$ and a point $A$, and we wish to know whether $A$ is within the acute 'angle' between $P$ and $Q$. Let $B$ be the point on $P$ such that $AB ⊥ P$. Let $C$ be the intersection of $AB$ and $Q$. Then we simply have to check whether $A$ is between $B$ and $C$ or not. If $A$ is rational and $P,Q$ are given by rational equations, $B$ and $C$ will also be rational and easy to find.

Another way (pointed out by DRF in chat) is that you can translate the point and the planes by the same vector so that the point goes to the origin, and then apply the test you already have for that case. Translation is rigid, so the answer before and after the translation is the same.

Answer 3

Every euclidian plane in 3D may be defined by a single point and an orthogonal vector to the plane. Both are not unique, but one may choose: $$5x-y+z+3=0 \Leftrightarrow (5, -1, 1) \cdot [(x, y, z) - (0, 0, -3)] = 0$$ $$4x-3y+2z+5=0 \Leftrightarrow (4, -3, 2) \cdot [(x, y, z) - (1, 3, 0)] = 0$$ Then, let us transpose everyone so $(3,2,-1)$ becomes the origin. We add $(-3, -2, 1)$ to the reference point in each plane. First plane becomes: $$(5, -1, 1) \cdot [(x, y, z) - (-3, -3, -2)] = 0 \Leftrightarrow 5x-y+z+14=0$$ And the second one: $$ (4, -3, 2) \cdot [(x, y, z) - (-2, 1, 1)] = 0 \Leftrightarrow 4x-3y+2z+9=0$$ Finally, apply the test: $$(5, -1, 1) \cdot (4,-3,2) = 25 >0$$ So, if I did not make any mistake, this point lies in the obtuse angle, since the product was positive.