June Revision Exercise 8 Q5

By KS Teng

(i)
l_{AB}: r = \begin{pmatrix}4 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix}1 \\ 2 \\ -1 \end{pmatrix}, \lambda \in \mathbb{R}

l_{AB}: r = \begin{pmatrix}7 \\ -3 \\ 13 \end{pmatrix} + \mu \begin{pmatrix}-2 \\ 1 \\ -5 \end{pmatrix}, \mu \in \mathbb{R}

If they intersect, then \begin{pmatrix}7 \\ -3 \\ 13 \end{pmatrix} + \mu \begin{pmatrix}-2 \\ 1 \\ -5 \end{pmatrix} = \begin{pmatrix}4 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix}1 \\ 2 \\ -1 \end{pmatrix} for some \lambda \text{ and } \mu

Solving with GC, we have \lambda = -1, \mu = 2. Thus, the lines intersect.

(ii)
\begin{pmatrix}-2 \\ 1 \\ -5 \end{pmatrix} \times \begin{pmatrix}1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix}-9 \\ 7 \\ 5 \end{pmatrix}

Equation of plane: r \bullet \begin{pmatrix}-9 \\ 7 \\ 5 \end{pmatrix} = \begin{pmatrix}4 \\ 1 \\ 2 \end{pmatrix} \bullet \begin{pmatrix}-9 \\ 7 \\ 5 \end{pmatrix} = -19

(iii)
Required distance

= \frac{|\vec{AE}\bullet \begin{pmatrix}-9 \\ 7 \\ 5 \end{pmatrix}|}{|\sqrt{9^2 + 7^2 + 5^2}|}

= \frac{3}{\sqrt{155}}

(iv)
Area of quadrilateral

= \text{ Area Triangle } ABC + \text{ Area Triangle } ACD

= \frac{1}{2}|\vec{CA} \times \vec{CD}| + \frac{1}{2}|\vec{AC} \times \vec{AB}|

= \frac{1}{2} |{\begin{pmatrix}-3 \\ 4 \\ -11 \end{pmatrix} \times \begin{pmatrix}-2 \\ 1 \\ -5 \end{pmatrix}}| + \frac{1}{2} |{\begin{pmatrix}3 \\ -4 \\ 1 \end{pmatrix} \times \begin{pmatrix}1 \\ 2 \\ -1 \end{pmatrix}}|

=\frac{1}{2} |\begin{pmatrix}-9 \\ 7 \\ 5 \end{pmatrix}| + \frac{1}{2}|\begin{pmatrix}-18 \\ 14 \\ 10 \end{pmatrix}|

= \frac{1}{2}(\sqrt{81 + 49 + 25} + 2\sqrt{81 + 49 + 25})

=\frac{3}{2}\sqrt{155}

Back to June Revision Exercise 8.

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