Find the Cartesian equation of the plane containing P, Q and R.

Given that П₁ and П₂ meet in a line $L$, verify that the vector equation of $L$ can be given by r $=\left(\begin{array}{c}\frac{5}{4}\\ 0\\ -\frac{7}{4}\end{array}\right)+\lambda \left(\begin{array}{c}\frac{1}{2}\\ 1\\ -\frac{5}{2}\end{array}\right)$.

Given that П₃ is parallel to the line $L$, show that $a+2b-5c=0$.

Show that $5a-7c=4$.

Given that П₃ is equally inclined to both П₁ and П₂, determine two distinct possible Cartesian equations for П₃.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

for example

$\overrightarrow{\text{PQ}}=\left(\begin{array}{c}-1\\ -5\\ 8\end{array}\right)$, $\overrightarrow{\text{PR}}=\left(\begin{array}{c}1\\ -6\\ 3\end{array}\right)$       A1A1

$\overrightarrow{\text{PQ}}\times \overrightarrow{\text{PR}}$ = 33i + 11j + 11k      (M1)A1

r.n = a.n

$33x+11y+11z=\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)\cdot \left(\begin{array}{c}33\\ 11\\ 11\end{array}\right)=22$     (M1)

$\Rightarrow 3x+y+z=2$ or equivalent       A1

METHOD 2

assume plane can be written as $ax+by+cz=1$       M1

substituting each set of coordinates gives the system of equations:

$a+6b-7c=1$

$0a+b+c=1$

$2a+0b-4c=1$       A1

solving by GDC      (M1)

$a=\frac{3}{2}$, $b=\frac{1}{2}$, $C=\frac{1}{2}$       A1A1A1

$\Rightarrow \frac{3}{2}x+\frac{1}{2}y+\frac{1}{2}z=1$ or equivalent

[6 marks]

METHOD 1

substitution of equation of line into both equations of planes       M1

$3\left(\frac{5}{4}+\frac{\lambda }{2}\right)+\left(-\frac{7}{4}-\frac{5\lambda }{2}\right)=2$       A1

$\left(\frac{5}{4}+\frac{\lambda }{2}\right)-3\lambda -\left(-\frac{7}{4}-\frac{5\lambda }{2}\right)=3$       A1

METHOD 2

adding Π₁ and Π₂ gives $4x-2y=5$       M1

given  $y=\lambda \Rightarrow x=\frac{5}{4}+\frac{\lambda }{2}$       A1

$z=2-y-3x=-\frac{7}{4}-\frac{5\lambda }{2}$       A1

⇒r $=\left(\begin{array}{c}\frac{5}{4}\\ 0\\ -\frac{7}{4}\end{array}\right)+\lambda \left(\begin{array}{c}\frac{1}{2}\\ 1\\ -\frac{5}{2}\end{array}\right)$       AG

METHOD 3

n₁ × n₂ = $\left(\begin{array}{c}2\\ 4\\ -10\end{array}\right)$       A1

$=4\left(\begin{array}{c}\frac{1}{2}\\ 1\\ -\frac{5}{2}\end{array}\right)$       R1

common point $\frac{5}{4}-3\left(0\right)-\left(-\frac{7}{4}\right)=3$ and $-3\left(\frac{5}{4}\right)-0-\left(-\frac{7}{4}\right)=-2$       A1 

[3 marks]

normal to П₃ is perpendicular to direction of $L$

$\Rightarrow \left(\begin{array}{c}a\\ b\\ c\end{array}\right)\cdot \left(\begin{array}{c}1\\ 2\\ -5\end{array}\right)=0$       A1

⇒$a+2b-5c=0$       AG

[1 mark]

substituting $\left(\begin{array}{c}\frac{5}{4}\\ 0\\ -\frac{7}{4}\end{array}\right)$ into П₃:      M1

$\frac{5a}{4}-\frac{7c}{4}=1$      A1

$5a-7c=4$      AG

[2 marks]

attempt to find scalar products for П₁ and П₃, П₂ and П₃.

and equating       M1

$\frac{3a+b+C}{\sqrt{11}\sqrt{a^2+b^2+c^2}}=\frac{a-3b-c}{\sqrt{11}\sqrt{a^2+b^2+c^2}}$      M1

Note: Accept $3a+b+c=a-3b-c$.

$\Rightarrow a+2b+c=0$      A1

attempt to solve $a+2b+c=0$, $a+2b-5c=0$, $5a-7c=4$      M1

$\Rightarrow a=\frac{4}{5}\text{,}b=-\frac{2}{5}\text{,}c=0$      A1

hence equation is $\frac{4x}{5}-\frac{2y}{5}=1$

for second equation:

$\frac{3a+b+C}{\sqrt{11}\sqrt{a^2+b^2+c^2}}=-\frac{a-3b-c}{\sqrt{11}\sqrt{a^2+b^2+c^2}}$      (M1)

$\Rightarrow 2a-b=0$

attempt to solve $2a-b=0$, $a+2b-5c=0$, $5a-7c=4$      M1

⇒$a=-2$, $b=-4$, $c=-2$      A1

hence equation is $-2x-4y-2z=1$

[7 marks]