IB Maths AA HL Ahl 3 17 Hl Planes Questions

(d) (i) METHOD 1
attempt to find cross product of two of $\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{BC}}$ or their opposites
eg $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left(\begin{array}{c}0 \\ k-9 \\ 18-2 k\end{array}\right)\left(=(k-9)\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)\right)$
attempt to substitute their cross product and a point into the equation of a plane
(k-9) y+2(9-k) z=2(k-9)+6(9-k)
$(k-9) y+2(9-k) z=36-4 k(\Rightarrow y-2 z=-4$ since $k \neq 9)$
METHOD 2
attempt to find vector equation of $\Pi$ and write x, y and z in parametric form
$\left(\boldsymbol{r}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)+\lambda\left(\begin{array}{c}k-1 \\ -4 \\ -2\end{array}\right)+\mu\left(\begin{array}{c}4 \\ -2 \\ -1\end{array}\right) \Rightarrow x=1+\lambda(k-1)+4 \mu, y=2-4 \lambda-2 \mu\right.$,
$z=3-2 \lambda-\mu$ or equivalent
attempt to eliminate both parameters to work towards Cartesian form
$(k-9) y+2(9-k) z=36-4 k(\Rightarrow y-2 z=-4$ since $k \neq 9)$

METHOD 1
attempt to find the equation of the line through (0,0,0) perpendicular to the plane

EITHER

attempt to find the point where the line and plane intersect

OR

attempt to find the point where the line and plane intersect

THEN
so the point on the plane closest to the origin is (0,-0.8,1.6)METHOD 2
choose a point on the plane (p, q, r)

distance to the origin is $\sqrt{p^{2}+(2 r-4)^{2}+r^{2}}$
since p is independent of r, distance is minimised when p=0
attempt to find the value of r for which their $\sqrt{(2 r-4)^{2}+r^{2}}$ is minimised

so the point on the plane closest to the origin is (0,-0.8,1.6)METHOD 3
attempt to find a vector from the origin to the closest point on the plane
EITHER
$(\boldsymbol{r}=) t\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)$
distance to the origin $=\left(\frac{4}{\sqrt{1^{2}+(-2)^{2}}}=\frac{4}{\sqrt{5}}\right)=\frac{4 \sqrt{5}}{5}$$t= \pm \frac{4}{5}$
check in equation of plane y-2 z=-4 to get $t=-\frac{4}{5}$
OR
$(\boldsymbol{r}=) t(k-9)\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)$
distance to the origin $=\left(\frac{4}{\sqrt{1^{2}+(-2)^{2}}}=\frac{4}{\sqrt{5}}\right)=\frac{4 \sqrt{5}}{5}$$t= \pm \frac{4}{5(k-9)}$
check in equation of plane y-2 z=-4 to get $t=-\frac{4}{5(k-9)}$
THEN
so the point on the plane closest to the origin is (0,-0.8,1.6)