Show that planes $x+2y+3z=8$ and $2x+3y+4z=11$ intersect in a line coplanar with $\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$

Question

Question is

Show that the line of intersection of the planes $$ x + 2y + 3z = 8 \quad\text{and}\quad 2x + 3y + 4z = 11 $$ is coplanar with the line $$\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$$

Also find the equation of the plane containing them.

$\text{Any hint how could I proceed ?}$

I know how to find the vector the line of intersection would be parallel to , which is given by cross product of the normal vectors of the two planes .

But that does not help in finding the equation of the line of intersection .

What I know is that to prove that two lines are coplanar we have to show that they intersect i.e the shortest distance between them is zero .

But unfortunately this formula again requires one point through which the line of intersection would pass which is unknown.

ps - I am in school yet :)

Answer 1

This might not be optimal, but here's a program you can follow:

1) Use the equations of your first and second lines to produce vectors which are contained in those lines.

2) Compute $\vec{n}$ the cross product of these vectors.

3) Using $\vec{n}$, find an equation of a plane containing one of the lines. Check to see that it contains both.

Let me know if you have trouble!

EDIT: I'll give you more information about 1). To get an equation of the first line, subtrace 2 times the first equation from the second. So we have $$\begin{array}{lll} &2x+3x+4z=11\\ -&2[x+2y+3z=8]\\\hline &0+-y-2z=-5 \end{array}$$ So, if you choose a $y$-coordinate, the $z-$coordinate of the point on the line with your chosen $y$ is $z=(-5+y)/-2$. Plugging this into equation 2, you get $2+3y+4[(-5+y)/-2]=11$ which gives $x=(21-5y)/2$. In particular, you notice that we have a parametrization for the line by $$y\mapsto (21/2-5/2y,y,5/2-2y)$$ with vector $\langle -5/2,1,-2\rangle$

Answer 2

One way of approaching this (again, not sure if this is optimal):

1) find the normal vector to each plane. The equations of the planes are given in the perfect form to easily determine this since one way of writing the equation of a plane is $\vec{r} \cdot \vec{n} = k$, where $k$ is a scalar. So the two normals are $(1,2,3)$ and $(2,3,4)$ (I keep forgetting how to represent column vectors in Mathjax, but assume that those are arrayed vertically).

2) You can now cross those two normals to find a vector that gives the direction of the line of intersection. It's easily computed using a determinant as $(2,2,-1)$ (again a column vector).

3) The equation of the line of intersection can now be found. You need one more piece of info - a single point on the line. Using the equations of the planes given, set $x=0$ and solve the remaining system to get $y=1, z=2$, giving one point on the line as $(0,1,2)$. You can now write down the equation of the line of intersection as $\vec r_1 = \lambda (2,2,-1) + (0,1,2)$.

4) Similarly, you can transform the equation of the other line given to $\vec r_2 = \mu(1,2,3) + (-1,-1,-1)$.

5) Now simply apply the condition for coplanarity of those lines by computing $\vec a = (0,1,2) - (-1,-1,-1) = (1,2,3)$ and $\vec b = (2,2,-1) \times (1,2,3) = (8,-7,2)$ and verifying that the dot product of those two is zero (it is indeed, as $8-14+6 = 0$). You can use this site as a reference for the final step: https://byjus.com/maths/coplanarity-two-lines/

Answer 3

You don’t need to find the line of intersection explicitly. Every plane that contains the line of intersection $L_1$ of the two given planes has an equation that’s a linear combination of their equations, i.e., of the form $\lambda(x+2y+3z-8)+\mu(2x+3y+4z-11)=0$. From the description of the second line $L_2$ we can immediately recover a point $P=(-1,-1,-1)$ on the line. If $L_2$ lies on any of these planes through $L_1$, then certainly $P$ must do so. This reduces the pool of candidates for the common plane of $L_1$ and $L_2$ to a single plane—the one define by $L_1$ and $P$. Substituting the coordinates of $P$ into the general equation above produces the equation $-14\lambda-20\mu=0$. I expect that you can come up a pair of values of the parameters that satisfy it. It remains to show that $L_2$ actually lies on this plane, which it does if its direction vector is perpendicular to the plane’s normal. Again, we can read a direction vector for $L_2$ directly from the description, namely, $(1,2,3)$ and I assume that you know how to extract a normal vector from the equation of a plane. Compute their dot product and see if it vanishes.

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Actually, you can avoid solving any equations at all and construct an equation for the common plane directly by using a technique sometimes called Plücker’s mu: The zero set of every linear combination $\lambda f+\mu g$ of two functions $f$ and $g$ contains the intersection of their individual zero sets. It’s easy to verify that a linear combination of $f$ and $g$ that also includes the point $\mathbf p$ in its zero set is $g(\mathbf p)f-f(\mathbf p)g$. Applying this to your problem, we’ve already constructed a generic linear combination of the equations of the two planes. Plücker’s mu tells us that $\lambda = 2(-1)+3(-1)+4(-1)-11 = -20$ and $\mu = -1(-1)-2(-1)-3(-1)+8 = 14$, consistent with the equation in the previous section. You should of course still verify that $L_2$ actually does lie on the resulting plane.

Answer 4

The line of intersection between the two planes can be represented parametrically. Let $z = t$:

$$ \left\{ \begin{array}{c} x + 2y = 8 - 3t \\ 2x + 3y = 11 - 4t \\ \end{array} \right. $$

Then $2 (1) - (2)$ gives $y = (16 - 6t) - (11 - 4t) = 5 - 2t$, and so $x = (8 - 3t) - 2(5 - 2t) = -2 + t$.

We have to check that these lines are not skew. This is a sufficient condition for two lines in 3D space to be coplanar, as two lines can at most span $\mathbb R^2$ (imagine rotating a plane around one line until it intersects the other line).

The other line can be rewritten parametrically as $(\lambda - 1, 2 \lambda - 1, 3 \lambda - 1)$. Thus we have:

$$ \left\{ \begin{array}{c} t - 2 = \lambda - 1 \\ -2t + 5 = 2 \lambda - 1 \\ t = 3 \lambda - 1 \\ \end{array} \right. $$

Substituting the third into the first, we have $(3 \lambda - 1) - 2 = \lambda - 1 \implies 3 \lambda - 3 = \lambda - 1$, thus $\lambda = 1$ and $t = 2$. Now we have to verify that $-2 \cdot 2 + 5 = 2 \cdot1 - 1$. This is true, so the lines are coplanar.

The equation of the plane is given by the normal vector to the direction vectors of the lines. This means we have to compute $(1, -2, 1)^T \times (1, 2, 3)^T$, which is $(-6-2, 1-3, 2--2)^T$ or $(-8, -2, 4)^T$. A multiple of this normal vector is $(4, 1, -2)^T$, so we need:

$$\begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} $$

or $4x + y - 2z = -3$.