IMO Practice Test — Three-Dimensional Geometry
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
hard
The image of the point $(1,3,4)$ in the plane $2x-y+z+3=0$ is:
$(-3,5,2)$
$(3,-1,6)$
$(-3,5,0)$
$(1,-1,4)$
Explanation: Normal direction: $(2,-1,1)$. Foot of perpendicular then reflection formula gives $(-3,5,2)$.
Question 2 of 6
hard
The angle bisector planes of $x+2y-2z=3$ and $2x-y+2z=1$:
$x+y=2$ and $x-5y+4z=0$
$3x+y=4$
Both A
$x+2y+z=4$
Explanation: $\frac{x+2y-2z-3}{3}=\pm\frac{2x-y+2z-1}{3}$. Adding/subtracting gives two angle bisector planes.
Question 3 of 6
hard
A plane meets the coordinate axes at $A$, $B$, $C$. If centroid of $\triangle ABC$ is $(1,2,3)$, the plane equation is:
$x+y/2+z/3=3$
$x/3+y/6+z/9=1$
$6x+3y+2z=18$
$x+2y+3z=18$
Explanation: If $A=(a,0,0)$, $B=(0,b,0)$, $C=(0,0,c)$: centroid $(a/3,b/3,c/3)=(1,2,3)$ gives $a=3,b=6,c=9$. Plane: $x/3+y/6+z/9=1$, i.e. $6x+3y+2z=18$.
Question 4 of 6
hard
The vector equation of the plane through three points $A(2,2,-1)$, $B(3,4,2)$, $C(7,0,6)$:
$5x+2y-3z=17$
$5x-2y+3z=17$
$5x+2y+3z=17$
$-5x+2y-3z=17$
Explanation: Normal $=(\vec{AB}\times\vec{AC})=(1,2,3)\times(5,-2,7)=(20,8,-12)$, simplified $(5,2,-3)$. Plane $5(x-2)+2(y-2)-3(z+1)=0\Rightarrow 5x+2y-3z=17$.
Question 5 of 6
medium
If a plane makes equal intercepts on all three axes, its equation is:
$x+y+z=0$
$x=y=z$
$x+y+z=c$ (where $c$ is a constant)
$xyz=k$
Explanation: Equal intercepts $a=b=c$. Plane: $x/a+y/a+z/a=1$, i.e. $x+y+z=a=c$ (constant).
Question 6 of 6
medium
A line passes through $(2,-1,3)$ and is perpendicular to $\vec{r}\cdot(2\hat{i}-\hat{j}+\hat{k})=4$. Its direction is:
$(2,-1,1)$
$(1,2,-1)$
$(-2,1,-1)$
$(2,1,-1)$
Explanation: The direction of a line perpendicular to a plane is along the plane's normal, $(2,-1,1)$.