All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)

Direction vector of l \text{~is~} \begin{pmatrix}{-3}\\6\\9\end{pmatrix} = -3 \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}

Normal vector of p \text{~is~} \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}

Since l is parallel to the normal vector of p, l is perpendicular to p.

(ii)
l: r = \begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}

[\begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}] \bullet \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix} = 0

10 + \lambda + 2 4 \lambda + 9 + 9 \lambda = 0

\lambda = - \frac{3}{2}

Point of intersection = (\frac{17}{2}, 2, \frac{3}{2}).

(iii)
\begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix} = \begin{pmatrix}{-2}\\23\\33\end{pmatrix}

10 + \lambda = - 2 — (1)

-1 - 2 \lambda = 23 — (2)

-3 - 3 \lambda = 33 — (3)

Since \lambda = -12 satisfies all 3 equations, A lies on l.

Since = (\frac{17}{2}, 2, \frac{3}{2}) is the midpoint of A & B, by ratio theorem,

\begin{pmatrix}{\frac{17}{2}}\\2\\{\frac{3}{2}}\end{pmatrix} = \frac{\vec{OA} + \vec{OB}}{2}

\vec{OB} = 2 \begin{pmatrix}{\frac{17}{2}}\\2\\{\frac{3}{2}}\end{pmatrix} - \begin{pmatrix}{-2}\\23\\33\end{pmatrix} = \begin{pmatrix}19\\{-19}\\{-30}\end{pmatrix}

B(19, -19, -30)

(iv)
Area

= \frac{1}{2} |\vec{OA} \times \vec{OB}|

= \frac{1}{2} |\begin{pmatrix}{-2}\\23\\33\end{pmatrix} \times \begin{pmatrix}19\\{-19}\\{-30}\end{pmatrix}|

= \frac{1}{2} |\begin{pmatrix}{-63}\\{567}\\{-399}\end{pmatrix}|

= \frac{1}{2} \sqrt{63^2 +567^2 + 399^2}

= 348 to the nearest whole number.

KS Comments:

Students must give answers in coordinates, rather than as a position vector.

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