# Find the image of the point 1,2,3 in the plane x + 2y + 4z - 38 =0 Help?

Jun 11, 2018

See below

#### Explanation:

The image of point $P = \left(1 , 2 , 3\right)$ has co-ordinates $Q = \left(\alpha , \beta , \gamma\right)$.

Midpoint of $P \text{ and } Q$ sits in plane $\boldsymbol{\pi} : x + \boldsymbol{2} y + \boldsymbol{4} z = 38$, so:

• $\boldsymbol{1} \frac{\alpha + 1}{2} + \boldsymbol{2} \frac{\beta + 2}{2} + \boldsymbol{4} \frac{\gamma + 3}{2} = 38$

$\therefore \alpha + 2 \beta + 4 \gamma = 59 q \quad \square$

And the line joining $P \text{ and } Q$ has same direction vector as the normal to the plane, $\boldsymbol{n} = \left(1 , 2 , 4\right)$, so:

$\boldsymbol{Q P} = k \setminus \boldsymbol{n}$

$\frac{\alpha - 1}{1} = \frac{\beta - 2}{2} = \frac{\gamma - 3}{4} = k$

• $\implies \left\{\begin{matrix}\beta = 2 \alpha \\ \gamma = 4 \alpha - 1\end{matrix}\right.$

Turn everything in $\square$ into $\alpha$:

$\alpha + 2 \left(2 \alpha\right) + 4 \left(4 \alpha - 1\right) = 59$

• $\implies \left\{\begin{matrix}\alpha = 3 \\ \beta = 6 \\ \gamma = 11\end{matrix}\right.$

Check:

$\boldsymbol{Q P} = \left(2 , 4 , 8\right) = 2 \setminus \boldsymbol{n} , q \quad \therefore k = 2$