1. ## equations planes

2. Originally Posted by Raiden_11
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Hello,

maybe this comes a little bit late ...
If $\vec{x}=\left(\begin{array}{c}x \\ y \\z \end{array} \right)$ then the equations of the lines can be written like this: $L:\vec{x} = \underbrace{\vec{a}}_{\text{fixed point}} + r \cdot \underbrace{\vec{u}}_{\text{direction vector}}$
$L_1: \vec{x} = \left(\begin{array}{c}5 \\ -1 \\1 \end{array} \right) + k \cdot \left(\begin{array}{c}3 \\ -4 \\0 \end{array} \right)$ and

$L_2: \vec{x} = \left(\begin{array}{c}1 \\ 1 \\1 \end{array} \right) + t \cdot \left(\begin{array}{c}1 \\ 2 \\-2 \end{array} \right)$

If $\vec{a}$ and $\vec{b}$ are the position vectors to the fixed points of the lines, $\vec{u}$ and $\vec{v}$ the direction vectors of the lines and $\vec{n}$ is the normal vector to $\vec{u}$ and $\vec{v}$ then the distance d is calculated by:

$d = \frac{(\vec{b} - \vec{a}) \cdot \vec{n}}{|\vec{n}|}$ (see attachment)

1. $\vec{n} = \left(\begin{array}{c}3 \\ -4 \\0 \end{array} \right) \times \left(\begin{array}{c}1 \\ 2 \\-2 \end{array} \right) = \left(\begin{array}{c}8 \\ 6 \\10 \end{array} \right)$ thus $\left| \left(\begin{array}{c}8 \\ 6 \\10 \end{array} \right) \right| = \sqrt{200} = 10 \cdot \sqrt{2}$

2. Now plug in all the values you know into the equation to calculate the distance:

$d = \frac{\left( \left(\begin{array}{c}1 \\ 1 \\1 \end{array} \right) - \left(\begin{array}{c}5 \\ -1 \\1 \end{array} \right) \right) \cdot \left(\begin{array}{c}8 \\ 6 \\10 \end{array} \right)}{10 \cdot \sqrt{2}} = \frac{-20}{10 \cdot \sqrt{2}} = -\sqrt{2}$

3. Originally Posted by Raiden_11

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Hello,

let $\overrightarrow{n_i}$ the normal vector of the plane $p_i$ that means $\overrightarrow{n_1}$ is :
$\overrightarrow{n_1} = \left( \begin{array}{c}a_1 \\ b_1 \\c_1 \end{array}\right)$

to a) $\overrightarrow{n_1} = k \cdot \overrightarrow{n_2} = t \cdot \overrightarrow{n_3}$ and $d_1 \ne d_2 \wedge d_1 \ne d_3 \wedge d_2 \ne d_3$

to b) $\overrightarrow{n_1} = k \cdot \overrightarrow{n_2} \text{ and } \overrightarrow{n_1} \ne \overrightarrow{n_3} \text{ and } \overrightarrow{n_2} \ne \overrightarrow{n_3}$

to c) $\overrightarrow{n_1} \ne \overrightarrow{n_2} \text{ and } \overrightarrow{n_3} = k \cdot \overrightarrow{n_1} + t \cdot \overrightarrow{n_2} \text{ and } d_3 = k \cdot d_1 + t \cdot d_2$

to d) $\overrightarrow{n_1} \times \overrightarrow{n_2} \text{ and } \overrightarrow{n_1} \times \overrightarrow{n_3} \text{ and } d_1 \ne d_2 \wedge d_1 \ne d_3 \wedge d_2 \ne d_3$

to e) That's the usual situation: No numerical dependencies(?) between the normal vectors and the constants.

4. ## what about the third one?

Thanks for the help earlier can u helpme solve this one too.

5. Originally Posted by Raiden_11
Thanks for the help earlier can u helpme solve this one too.

Hello,

to a): I assume that you have written your problem correctly.
If so you can simplify the given equations:
$x = (2-k)(4)-3k = 8-7k$
$y = (2-k)(-6) = -12 + 6k$
$z = (2-k)+2k = 2 + k$

If $\vec{x}=\left(\begin{array}{c}x\\y\\z \end{array}\right)$

Then these equations with the given restriction describe a straight line connecting the points A(8, -12, 2) [for k = 0] and B(-6, 0, 4) [for k = 2]:

The equation of the line where the points A and B are placed on is:

$\vec{x}=\left(\begin{array}{c}8\\ -12 \\ 2 \end{array}\right) + k \cdot \left(\begin{array}{c}-7 \\ 6\\ 1 \end{array}\right)$

to b) I have no idea what to do - sorry!