Three-dimensional geometry: lines and planes practice questions — IB Maths AA

The tetrahedron with these vertices has volume $1\mathrm{/}6\mathrm{\ne }0$, so the four points are NOT coplanar — they form a non-degenerate tetrahedron.

The three edges from origin are $(3,0,0),(0,4,0),(0,0,5)$. Determinant: $3\cdot 4\cdot 5=60$. Volume: $60\mathrm{/}6=10$.

$3a-1-2=0\Rightarrow 3a=3\Rightarrow a=1$.

$r^2=49\Rightarrow r=7$.

$d=\mathrm{\mid }-6\mathrm{\mid }\mathrm{/}\sqrt{4+4+1}=6\mathrm{/}3=2$.

A general point of $\mathrm{\ell }$ has coordinates

$$(x,y,z)=(1+2t,2-t,-1+3t)\mathrm{.}$$

Such a point lies on $\pi$ exactly when its coordinates satisfy $x+y+z=10$:

$$(1+2t)+(2-t)+(-1+3t)=10.$$

Collecting the constants and the terms in $t$:

$$2+4t=10⟹4t=8⟹t=2.$$

Substituting $t=2$ back into the parametric coordinates:

$$x=1+2(2)=5,y=2-2=0,z=-1+3(2)=5.$$

The check $5+0+5=10$ confirms the point lies on $\pi$, and $t=2$ places it on $\mathrm{\ell }$. Since the direction vector $(2,-1,3)$ is not perpendicular to the normal $(1,1,1)$ — their dot product is $2-1+3=4\mathrm{\ne }0$ — the line is not parallel to the plane, so this intersection point is unique:

$$(5,0,5)\mathrm{.}$$

$(1,0,0)\times (0,2,0)=(0\cdot 0-0\cdot 2,0\cdot 0-1\cdot 0,1\cdot 2-0\cdot 0)=(0,0,2)$.

Locate the endpoint of the median, the midpoint $M$ of $BC$ with $B(5,0,-1)$ and $C(-1,4,3)$: $$M=(\frac{5+(-1)}{2},\frac{0+4}{2},\frac{-1+3}{2})=(2,2,1)$$

Take the coordinate differences between $A(1,2,3)$ and $M(2,2,1)$: $$\mathrm{\Delta }x=2-1=1,\mathrm{\Delta }y=2-2=0,\mathrm{\Delta }z=1-3=-2$$

Square, add, and take the square root: $$\mathrm{\mid }AM\mathrm{\mid }=\sqrt{1^2+0^2+(-2{)}^2}=\sqrt{1+0+4}=\sqrt{5}$$

$d=\sqrt{3^2+4^2+0^2}=\sqrt{25}=5$. (3-4-5 triple in the $xy$-plane.)

Average each coordinate: $((2+8)\mathrm{/}2,(-1+5)\mathrm{/}2,(4-2)\mathrm{/}2)=(5,2,1)$.