Exercise 4 · Option B · Geometry in space
Equation of a plane through three points, area of the triangle with the cross product and the condition for a fourth point to form a tetrahedron of volume 1.
Maximum score · 2.5 pointsIn the exam you must choose one of the two options (A or B). Here you have Option B worked out; Option A is on the previous page.
Consider the points in space $P=(1,0,-1)$, $Q=(3,-2,0)$ and $R=(1,1,1)$.
- Compute the equation of the plane containing the points $P$, $Q$ and $R$. 0.75 p
- Verify that the area of the triangle $\triangle PQR$ is $\dfrac{3\sqrt{5}}{2}$. 0.75 p
- Find the conditions that the coordinates of a fourth point $S=(x,y,z)$ must satisfy so that $P$, $Q$, $R$ and $S$ form a tetrahedron of volume 1. (The volume of the tetrahedron formed by the points $P,Q,R,S$ is $V=\tfrac{1}{6}\,\big|(\vec{PQ}\times\vec{PR})\cdot\vec{PS}\big|$.) 1 p
Step-by-step solution
Key idea: two vectors of the plane, $\vec{PQ}$ and $\vec{PR}$, control everything. Their cross product gives the normal vector (for the plane) and its magnitude (for the area); the scalar triple product with $\vec{PS}$ gives the volume of the tetrahedron.
4.1) Equation of the plane through $P$, $Q$, $R$
We compute two direction vectors of the plane:
The normal vector is the cross product:
The plane passes through $P=(1,0,-1)$, so $-5(x-1)-4(y-0)+2(z+1)=0$. Expanding (and changing sign):
4.2) Area of the triangle $\triangle PQR$
The area of the triangle is half the magnitude of the cross product:
4.3) Condition for the tetrahedron to have volume 1
Let $\vec{PS}=(x-1,\,y,\,z+1)$. The scalar triple product is the dot product of the normal with $\vec{PS}$:
We impose that the volume of the tetrahedron is 1:
The absolute value gives two cases:
These are two planes parallel to the plane $\pi$ (same normal vector), one on each side, located at the distance that makes the tetrahedron have volume 1.