# Find the equation of the plane that contains

Find the equation of the plane that contains,

the lines: $$\frac{x-2}{2} = \frac{y+4}{3} = \frac{2 - z}{5}$$ and \begin{align} x &= 3 + 4t \\ y &= -4 +6t \\ z &= 5 -10t. \end{align}

I'm not exactly sure on how to tackle this problem.

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Do you have any ideas at all about how to start this? And: If this is homework, then please add the homework tag. –  Thomas Oct 9 '12 at 2:15
Not homework, studying for an exam. –  Mike Oct 9 '12 at 2:29

The line $\frac{x-2}{2} = \frac{y+4}{3} = \frac{2-z}{5}\,$ can be written as $$x=2+2t\,,\quad y=-4+3t \,,\quad z=2-5t \,,$$ and the line $$x = 3 + 4t \,,y = -4 + 6t \,,z = 5 - 10t \,.$$

The two lines have the same direction, since $v_1=(2,3,-5)$ and $v_2 = (4, 6,-10)\,,$ where $v_1$ and $v_2$ are the direction vectors of the two lines.

One can get two points lies in the plane. Putting $t=0$ in the equations of the lines gives $p_1=(2,-4,2)$ and $p_2=(3,-4,5)\,,$ which lie in the plane.

Constructing the vector $v_3=p_2-p_1$ gives $v_3=(1,0,3)\,.$ Now, we can find the normal to the plane by taking the cross product of $v_3$ and $v_1$ or $v_2$.

$$n = v_3 \times v_2 \,.$$

Once that done, the equation of the plane is given by

$$n.(X-p_1)=0 \Rightarrow n.( x-2,y-4,z-2 )=0 \,.$$

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So this should give me, 31 x-11 y-19 z+20 = 0? –  Mike Oct 9 '12 at 2:49
@Mike:I did not carry on the calculations, but basically this is the techniques. –  Mhenni Benghorbal Oct 9 '12 at 2:51
The reason I ask is because the posted solution to this problem is 9x-11y-3z=56 –  Mike Oct 9 '12 at 2:52
@Mike: I made some corrections to the names of the vectors, make sure you have a look at that. –  Mhenni Benghorbal Oct 9 '12 at 2:55
In doing that I get <31,11,19> –  Mike Oct 9 '12 at 3:02

Hints: To find a plane all you need is a point and a normal vector. You can get one point from one line.

The normal vector $n$ needs to be perpendicular on both lines, thus perpendicular on both their directions. How do you find a vector perpendicular on two given directions?

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Hint: Note that the two lines are are parallel - both with direction vector $v = (2, 3, -5)$. Now for an equation of a plane you need a point and normal vector. The point should be easy to find.

For the normal vector, pick two point on each of the two lines given, say points $A$ and $B$. Then (try to draw a picture) a normal vector will be $v\times \vec{AB}$.

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