Equal spacing between parallel planes · source reference

Printed question

The plane $p_1$ has equation $x+y-2z=4$. A plane $p_2$ with equation $2x+ay+bz=-4$ (where $a,b$ are constants) is parallel to $p_1$, find the values of $a,b$ and also the exact distance between $p_1,p_2$. [3]

The plane $p_3$ is parallel to $p_1,p_2$. Find the equation of $p_3$ such that $p_2$ is exactly in between $p_1,p_3$. [2]

[modified 2010/NYJC/Prelim/II/4]

Printed working

$p_1\parallel p_2\Rightarrow(2,a,b)=k(1,1,-2)\Rightarrow k=2,a=2,b=-4$.

$p_1:\mathbf r\cdot(1,1,-2)=4\Rightarrow\mathbf r\cdot\frac{(1,1,-2)}{\sqrt6}=\frac4{\sqrt6}$.

$p_2:\mathbf r\cdot(2,2,-4)=-4\Rightarrow\mathbf r\cdot(1,1,-2)=-2\Rightarrow\mathbf r\cdot\frac{(1,1,-2)}{\sqrt6}=-\frac2{\sqrt6}$.

Perpendicular distance between $p_1,p_2=\frac4{\sqrt6}-\frac{-2}{\sqrt6}=\frac6{\sqrt6}=\sqrt6$.

From diagram, $p_3$ is on the same side of the origin $O$ as $p_2$, and the perpendicular distance from $O$ is $\frac2{\sqrt6}+\frac6{\sqrt6}=\frac8{\sqrt6}$.

Thus $p_3:\mathbf r\cdot\frac{(1,1,-2)}{\sqrt6}=-\frac8{\sqrt6}\Rightarrow\mathbf r\cdot(1,1,-2)=-8$.

Teacher notes

$p_1\parallel p_2\Rightarrow\mathbf n_1\parallel\mathbf n_2$.

Make the 2 parallel normal vectors of the 2 planes the same and of magnitude 1, i.e. unit vectors.

Printed answer key

$a=2,b=-4$; $\sqrt6$ units; $\mathbf r\cdot(1,1,-2)=-8$.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question booklet PDF pages 4–12 (printed 80–88); solution booklet PDF pages 17–44. Column vectors are typeset as ordered coordinate triples here; the entries, products, intermediate equations and original parameter/domain restrictions are retained.