Animated Solution for Mathematics - Three Dimensional Geometry: A plane passing through the points $(0,-1,0)$ and $(0,0,1)$ and making an angle $\pi /4$ with the plane $y-z+5=0$, also passes through the point

Imagine you are standing in a vast, empty three-dimensional coordinate system. You have two points, $P_1(0,-1,0)$ and $P_2(0,0,1)$, floating in the void.

You are tasked with creating a flat, infinite sheet—a plane—that passes through both of these points. This plane rotates like a door on a hinge, where the line connecting $P_1$ and $P_2$ acts as the axis of rotation.

Our mission is to find the specific orientation of this plane such that it makes an angle of $\frac{\pi }{4}$ (or $45^{\circ }$) with the fixed plane $y-z+5=0$.

Every plane in 3D space has a unique 'DNA' represented by the general equation $ax+by+cz+d=0$. The coefficients $(a,b,c)$ define the normal vector $\vec{n}$, which acts as a compass pointing directly away from the surface.

By substituting our points $P_1(0,-1,0)$ and $P_2(0,0,1)$ into this equation, we 'pin' the plane to those coordinates:

Substituting these into the general form, we get $ax+dy-dz+d=0$. Assuming $d eq 0$ and defining $k=\frac{a}{d}$, we arrive at the elegant form:

The fixed plane $y-z+5=0$ has a normal vector ${\vec{n}}_2=(0,1,-1)$. Our plane's normal vector is ${\vec{n}}_1=(k,1,-1)$.

The angle $\theta$ between the planes is the angle between these two normal vectors, determined by the dot product formula:

Calculating the components, we find the dot product is $(k)(0)+(1)(1)+(-1)(-1)=2$. The magnitudes are $|{\vec{n}}_1|=\sqrt{k^2+2}$ and $|{\vec{n}}_2|=\sqrt{2}$.

Setting $\cos (45^{\circ })=\frac{1}{\sqrt{2}}$, we establish the following relationship:

The $\sqrt{2}$ terms cancel out, leaving $1=\frac{2}{\sqrt{k^2+2}}$. Squaring both sides yields $k^2+2=4$, which simplifies to $k^2=2$.

Thus, we find $k=\pm \sqrt{2}$. The two possible planes satisfying the condition are:

You have successfully navigated the 3D rotation and constrained the plane to its final, elegant position. The resulting equations are $\sqrt{2}x+y-z+1=0$ and $-\sqrt{2}x+y-z+1=0$.