I'm facing some problems by finding the one missing point on plane $E$. We want to find $O(x,y,z)$ by using two known points $P(2,3,5)$ and $Q(1,2,2)$ and angle of $POQ = 60$ degrees with plane $E: 1.15714286x + 1.8547619y - z -2.86101191 = 0$ and length $OP = OQ$.
Is there any way to find it?
Thank you.
On
If you proceed step by step it is not that hard. From the nature of point $O$, it would lie on the perpendicular bisector of the line $PQ$, which lies in the plane E.
Let us first find this line. For that, we need to know a point through which it passes, and it's direction
Point - It passes through the midpoint of PQ - which can be found easily (point $\vec{M}$)
Direction - it is a vector perpendicular to the plane normal as well as the direction of line PQ. Hence take the cross product of both of these vector to get the vector of the line $\hat{l}$
Now that we have these two, we can write any point on the line
$$\vec{A} = \vec{M} + \lambda\hat{l}$$
Since the given triangle is iscoceles, you can directly obtain $\lambda$ using some trigonometry with the given angles to find what the length of $\vec{EM}$ would be
You have 3 conditions:
1). point $O(x,yz)$ lies on plane $E$, so $$ 1.15714286x + 1.8547619y - z =2.86101191$$
2). $OP=OQ$ $$(x-2)^2+(y-3)^2+(z-5)^2=(x-1)^2+(y-2)^2+(z-2)^2.$$ This simplifies to $$2x+2y+6z=29.$$
3). $\angle POQ=60^{\circ}$ and $OP=OQ$ so $\triangle POQ$ is equilateral. Thus
$$OQ^2=PQ^2 \implies (x-1)^2+(y-2)^2+(z-2)^2=11.$$
The answer will not be unique.