2022 DHS J2 MYE P1 Q10

2022 DHS J2 MYE P1 Q10

11 marks
Mid Year Exam

The plane \({{\pi }_{1}}\) contains the point \(A\left( 2,1,-1 \right)\) and the line \(l\) with equation \(\frac{x-5}{2}=y\), \(z=1\). Another plane \({{\pi }_{2}}\) meets \({{\pi }_{1}}\) in the line \(l\). It is known that \(\left( \begin{matrix} -1 \\ 2 \\ -6 \\ \end{matrix} \right)\) is a normal for \({{\pi }_{2}}\).

  1. Show that the equation of \({{\pi }_{1}}\) in scalar product form is \(\mathbf{r}.\left( \begin{matrix} -2 \\ 4 \\ 5 \\ \end{matrix} \right)=-5\).[3]
  2. Find the position vector of the foot of perpendicular, \(F\) from \(A\) to the line \(l\).
    Hence or otherwise, find the perpendicular distance from \(A\) to \({{\pi }_{2}}\).[3]
  3. Find the acute angle between \({{\pi }_{1}}\) and \({{\pi }_{2}}\).[2]
  4. \(A’\) is the point of reflection of \(A\) about the plane \({{\pi }_{2}}\). Deduce the area of the triangle \(AFA’\), leaving your answer in the form \(\frac{a\sqrt{5}}{b}\), where \(a\) and \(b\) are integers to be determined.[3]

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Answer:(a) \(\mathbf r\mathbin{\cdot}\begin{pmatrix}-2\\4\\5\end{pmatrix}=-5\) (b) \(\overrightarrow{OF}=\begin{pmatrix}3\\-1\\1\end{pmatrix}\), distance \(=\dfrac{17}{\sqrt{41}}\text{ units}\) (c) \(62.2^\circ\) (d) \(\dfrac{68\sqrt5}{41}\text{ units}^2\)

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