June Revision Exercise 8 Q6

By KS Teng

(a)
\vec{OC}

= \frac{\vec{OA}+3\vec{OB}}{4}

= \frac{1}{4} [{\begin{pmatrix}1 \\ 1 \\ 2 \end{pmatrix} + \begin{pmatrix}3 \\ 5 \\ 6 \end{pmatrix}}]

= \begin{pmatrix}2.5 \\ 4 \\ 5 \end{pmatrix}

X \text{lies in } \pi: \begin{pmatrix}2.5 \\ 4 \\ 5 \end{pmatrix} \bullet \begin{pmatrix}2 \\ \lambda \\ 0 \end{pmatrix}= \mu

5 + 4 \lambda = \mu

(b)
\vec{AB} = \begin{pmatrix}3 \\ 5 \\ 6 \end{pmatrix} - \begin{pmatrix}1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix}2 \\ 4 \\ 4 \end{pmatrix}

If the plane \pi contains the line AB, the p_3 is parallel to AB:

\begin{pmatrix}2 \\ 4 \\ 4 \end{pmatrix} \bullet \begin{pmatrix}2 \\ \lambda \\ 0 \end{pmatrix} = 0

4 + 4 \lambda = 0

\lambda = -1

\lambda = -1 and the point A lies in the plane p_3:

\begin{pmatrix}1 \\ 1 \\ 2 \end{pmatrix} \bullet \begin{pmatrix}2 \\ -1\\ 0 \end{pmatrix} = \mu

2 - 1 = \mu

\mu = 1

Alternatively, you may consider that A \text{and} B \text{line in the plane} p_3:

\begin{pmatrix}1 \\ 1 \\ 2 \end{pmatrix} \bullet \begin{pmatrix}2 \\ \lambda \\ 0 \end{pmatrix} = \mu

\Rightarrow 2 + \lambda = \mu — (1)

\begin{pmatrix}3 \\ 5 \\ 6 \end{pmatrix} \bullet \begin{pmatrix}2 \\ \lambda \\ 0 \end{pmatrix} = \mu

\Rightarrow 6 + 5\lambda = \mu — (2)

Solving, \lambda = -1, \mu = 1

Back to June Revision Exercise 8.

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    June Revision Exercise 8JC Mathematics
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