Vector Equation of a plane

Question

The lines $L1$ and $L2$ have equations $r = 8i - 14j + 13k + s (-4i + 7j - 6k)$ and $x/2 = (y-17)/5 = (z+7)/-1$ respectively. The plane contains both $L1$ and $L2$:

a) Find the vector equation of the plane:

Obviously, the direction vectors of the plane are simply the direction vectors of the lines, although there are a number of different solutions to the direction vector I will be using these, that is $d1 = -4i + 7j - 6k$ and $d2 = 2i + 5j - k$.

So $x/2 = (y-17)/5 = (z+7)/-1$ is the equivalent of $r = (17j - 7k) + t (2i + 5j - k)$.

I am unsure how to find a point that I know the plane will pass through, it could be the interception of $L1$ and $L2$, but in that case by finding the intercept you get $s = 76/17$ and $t = 1/17$ as the solutions from comparing the $x$ and $y$ components, which when you plug back in gets a horrendous solution.

The textbook claims the answer is $(-4, 7, -5)$, but where do they get this point from?

Answer 1

You’ve clearly made some arithmetic or algebraic error in computing the intersection of the two lines. Setting $s=76/17$ in $L_1$ produces $\frac1{17}(-168,294,-235)$, while setting $t=1/17$ in your parametric equation for $L_2$ produces $\frac1{17}(-2,294,-120)$.

From the two parametric equations we get the system $$\begin{align} -4s+8 &= 2t \\ 7s-14 &= 5t+17 \\ -6s+13 &= -t-7,\end{align}$$ which we can rearrange into $$\begin{align} -4s-2t &= -8 \\ 7s-5t &= 31 \\ -6s+t & = -20, \end{align}$$ with solution $s=3$, $t=-2$. If you plug these values into the parametric equations of the lines, you’ll find that their intersection is in fact $(-4,7,-5)$.

Of course, we needn’t have gone to all of that trouble. Since both lines lie entirely on the plane, any point on either line will do, such as $(8,-14,13)$ or $(0,17,-7)$, both of which can be read directly from the original equations. On the other hand, it’s good to verify that the lines do in fact intersect.

Answer 2

The lines are given in the following form

$$ \begin{aligned} \boldsymbol{L}_1 & = \boldsymbol{r}_1 + s\, \boldsymbol{e}_1 = \pmatrix{8 \\ -14 \\ 13} + s\, \pmatrix{-4 \\ 7 \\ -6} \\ \boldsymbol{L}_2 & = \boldsymbol{r}_2 + t\, \boldsymbol{e}_2 = \pmatrix{0 \\ 17 \\ -7} + t\, \pmatrix{2 \\ 5 \\ -1} \end{aligned} $$

_{Hint: Plug $\boldsymbol{L_2} = \pmatrix{x & y & z}$ into the second line expression to see that the above is an alternate form of the same line.}

To find the plane that both lines go through, you need to find the direction that is normal the plane. This is defined by the vector cross product of the line directions.

$$ \boldsymbol{n} = \boldsymbol{e}_1 \times \boldsymbol{e}_2 = \pmatrix{-4 \\ 7 \\ -6} \times \pmatrix{2 \\ 5 \\ -1} = \pmatrix{23 \\ -16 \\ -34} $$

Now any point on the plane projected along the normal should produce the same value.

$$ \boldsymbol{n} \cdot \pmatrix{x \\ y \\ z} = \text{const.} $$

This constant can be found by projecting any point we know belongs to the plane, such as $\boldsymbol{r}_1$ since all points of L1 or L2 should belong to the plane.

$$ \boldsymbol{n} \cdot \pmatrix{x \\ y \\ z} = \boldsymbol{n} \cdot \pmatrix{8 \\ -14 \\ 14} $$ $$ 23 x - 16 y - 34 z = -34 $$

In vector form the above is

$$ \pmatrix{23 \\ -16 \\ -34} \cdot \boldsymbol{r} = -34 $$ where $\boldsymbol{r} = \pmatrix{x & y & z} $.

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Proof

Show that

$$ \begin{aligned} \pmatrix{23 \\ -16 \\ -34} \cdot \boldsymbol{L}_1 & = -34 & \text{and}\\ \pmatrix{23 \\ -16 \\ -34} \cdot \boldsymbol{L}_2 & = -34 \end{aligned} $$