(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.

### One Comment

• […] June Revision Exercise 8 Q6 – The Culture on June Revision Set 8 […]