3D Coordinate Geometry practice questions — A-Level Maths

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

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: $$\mathrm{\mid }AM\mathrm{\mid }=\sqrt{1^2+0^2+(-2{)}^2}=\sqrt{1+0+4}=\sqrt{5}$$

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

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:

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

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

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

$1\cdot 2+2\cdot (-1)+3\cdot 1=2-2+3=3$.

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

Set up one equation per coordinate from $A(2,-4,6)$ and $M(4,0,2)$: $$\frac{2+x_B}{2}=4,\frac{-4+y_B}{2}=0,\frac{6+z_B}{2}=2$$

Multiply each equation by $2$: $$2+x_B=8,-4+y_B=0,6+z_B=4$$

Solve each one: $$x_B=6,y_B=4,z_B=-2$$

Check: the midpoint of $A(2,-4,6)$ and $(6,4,-2)$ is $({\displaystyle \frac{2+6}{2}},{\displaystyle \frac{-4+4}{2}},{\displaystyle \frac{6-2}{2}})=(4,0,2)=M$. $$B(6,4,-2)$$

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