IB Maths AA HL 3.17 Planes Question Bank

Practise IB Mathematics HL 3.17 by applying planes methods to exam-style questions.

[Maximum number: 9]

Consider the points A(1,2,3), B(k,-2,1) and C(5,0,2), where $k \in \mathbb{R}$ .

(a)

For $k \neq 9$ , let $\Pi$ be the plane containing A, B and C .

[ 9 ]

(i)

Find the Cartesian equation of the plane $\Pi$ .

(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 $k \neq 9)$

(ii)

Find the coordinates of the point on the plane $\Pi$ which is closest to the origin (0,0,0).

[ 9 ]

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

EITHER

(r=) t\left(\begin{array}{c} 0 \\ 1 \\ -2 \end{array}\right)

attempt to find the point where the line and plane intersect

t+4 t+4=0

t=-\frac{4}{5}

OR

(r=) t(k-9)\left(\begin{array}{c} 0 \\ 1 \\ -2 \end{array}\right)

attempt to find the point where the line and plane intersect

\begin{aligned} & t(k-9)^{2}+4 t(k-9)^{2}+4(k-9)=0 \\ & t=-\frac{4}{5(k-9)} \end{aligned}

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)

q-2 r+4=0 \text { OR } q(k-9)-2 r(k-9)+4(k-9)=0 \Rightarrow q=2 r-4

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

r=1.6

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
$(\boldsymbol{r}=) t\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)$
(A1)
distance to the origin $=\left(\frac{4}{\sqrt{1^{2}+(-2)^{2}}}=\frac{4}{\sqrt{5}}\right)=\frac{4 \sqrt{5}}{5}$
(A1)
$t= \pm \frac{4}{5}$
check in equation of plane y-2 z=-4 to get $t=-\frac{4}{5}$
OR
$(\boldsymbol{r}=) t(k-9)\left(\begin{array}{c}0 \\ 1 \\ -2\end{array}\right)$
(A1)
distance to the origin $=\left(\frac{4}{\sqrt{1^{2}+(-2)^{2}}}=\frac{4}{\sqrt{5}}\right)=\frac{4 \sqrt{5}}{5}$
(A1)
$t= \pm \frac{4}{5(k-9)}$
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)

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AHL 3.17 (HL)—Planes question 1

Question (a)

Question (i)

Question (ii)