Condition for coplanarity of two lines

[Aurelius120]

Homework Statement

Find the value of K for which the given lines are coplanar

Relevant Equations

NA

So I tried to solve it this way:
The equations of the lines in vector form are
$$(x-2)\hat i+(y-3)\hat j+(z-4)\hat k=\lambda (\hat i+\hat j-K\hat k)$$
$$(x-1)\hat i+(y-4)\hat j+(z-5)\hat k=\mu (K\hat i+2\hat j+1\hat k)$$

Since the lines are some real multiple of the vectors,
For coplanarity $$(\hat i+\hat j-K\hat k)\times (K\hat i+2\hat j+1\hat k)=0$$
Therefore, $2-k=0; -k^2-1=0; 1+2k=0$
So no solutions should exist, right?

But the book and some websites solve it thus

So where and why did I go wrong??

 

[pasmith]

Yhe only way the cross product of the tangent vectors will vanish is if the lines are parallel, which is not what was asked for.

 

[renormalize]

Yhe only way the cross product of the tangent vectors will vanish is if the lines are parallel, which is not what was asked for.

If two lines lie in the same plane, shouldn't they either intersect or be parallel?

 

[pasmith]

Homework Statement: Find the value of K for which the given lines are coplanar
Relevant Equations: NA

View attachment 343125
So I tried to solve it this way:
The equations of the lines in vector form are
$$(x-2)\hat i+(y-3)\hat j+(z-4)\hat k=\lambda (\hat i+\hat j-K\hat k)$$
$$(x-1)\hat i+(y-4)\hat j+(z-5)\hat k=\mu (K\hat i+2\hat j+1\hat k)$$

Since the lines are some real multiple of the vectors,
For coplanarity $$(\hat i+\hat j-K\hat k)\times (K\hat i+2\hat j+1\hat k)=0$$
Therefore, $2-k=0; -k^2-1=0; 1+2k=0$
So no solutions should exist, right?

But the book and some websites solve it thus
View attachment 343126
So where and why did I go wrong??

If \mathbf{x_0} + \lambda\mathbf{n_0} and \mathbf{x_!} + \mu\mathbf{n_1} lie in the same plane, then the normal to that plane must be \mathbf{n}_0 \times \mathbf{n}_1. However, a line drawn between a point on one line and a point on the other - such as the line from \mathbf{x}_0 to \mathbf{x}_1 - must also lie in that plane, so we must have <br /> (\mathbf{n}_0 \times \mathbf{n}_1) \cdot (\mathbf{x}_1 - \mathbf{x}_0) = 0. The determinant in the given answer calculates this triple scalar product.

If two lines lie in the same plane, shouldn't they either intersect or be parallel?

In 3D, there is also the possibility that the lines are not parallel and do not intersect, in which case they are not coplanar. If they intersect or are parallel, then indeed they are coplanar. Here the OP has shown that the lines cannot, in fact, be parallel - but depending on k they might not intersect.

 

[Gavran]

must have (n0×n1)⋅(x1−x0)=0.

The condition for coplanarity of two lines

$\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$

can be written as a triple product and in the next form

$((x_2-x_1)\hat i + (y_2-y_1)\hat j + (z_2-z_1)\hat k) \cdot ((a_1\hat i + b_1\hat j + c_1\hat k) \times (a_2\hat i + b_2\hat j + c_2\hat k)) = 0$.

 

[WWGD]

The condition for coplanarity of two lines
$\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$

can be written as a triple product and in the next form

$((x_2-x_1)\hat i + (y_2-y_1)\hat j + (z_2-z_1)\hat k) \cdot ((a_1\hat i + b_1\hat j + c_1\hat k) \times (a_2\hat i + b_2\hat j + c_2\hat k)) = 0$.

Moderator note: In a later post, WWGD recognizes that the following is in error.
Not sure I understand you. Unless the plane is given in advance, any two lines determine a plane by defining their cross product; the cross product of vectors along the respective lines.

 

[pasmith]

Not sure I understand you. Unless the plane is given in advance, any two lines determine a plane by defining their cross product; the cross product of vectors along the respective lines.

What plane is defined by (1) the x axis, and (2) the line parallel to the y axis and lying in the plane z = 1?

Two lines do not define a plane unless they are parallel or intersect. If that condition is not met, the lines do not lie in a common plane.

 

[WWGD]

What plane is defined by (1) the x axis, and (2) the line parallel to the y axis and lying in the plane z = 1?

Two lines do not define a plane unless they are parallel or intersect. If that condition is not met, the lines do not lie in a common plane.

Moderator note: In a later post, WWGD recognizes that the following is in error.
They don't define one directly, but they're in the plane whose normal is given by the cross-product of line segments in each. Your condition of being in the line $z=1$ is extraneous, as it's not what I claimed.

My claim is: Given any 2 lines $L_1, L_2$ in $\mathbb R^3$, there is a plane $P$ containing $L_1, L_2$.
Proof:
Take finite segments $v_1,v_2$ along $L_1, L_2$ respectively. Then the plane $P_{12}$, whose normal is the cross product $\pm |v_1 \times v_2|$, passing through any point $p$ in either line, contains both lines $L_1, L_2$.

 

[WWGD]

That's why I stated that, unless the plane is specified beforehand, the problem is ill-posed, as , given any two lines, there is always a plane $P$ containing them.
But maybe I misunderstood, misread the problem.

 

[WWGD]

Moderator note: In a later post, WWGD recognizes that the following is in error.
@FactChecker , not sure what you're doubtful about. Isn't the vector $\pm v_1,v_2$ perpendicular to both line segments, thus the two lines $L_1, L_2$? Doesn't it define a normal vector $N_{12}$ to a plane containing both, using any point $p_i; i=1,2$, so that

$N_{12}(p-p_i)=0$,

The equation of a plane containing both lines?

 

[FactChecker]

They don't define one directly, but they're in the plane whose normal is given by the cross-product of line segments in each. Your condition of being in the line $z=1$ is extraneous, as it's not what I claimed.

My claim is: Given any 2 lines $L_1, L_2$ in $\mathbb R^3$, there is a plane $P$ containing $L_1, L_2$.
Proof:
Take finite segments $v_1,v_2$ along $L_1, L_2$ respectively. Then the plane $P_{12}$, whose normal is the cross product $\pm |v_1 \times v_2|$, passing through any point $p$ in either line, contains both lines $L_1, L_2$.

I am not convinced of your proof. Can you show me that there is a plane in $\mathbb{R}^3$ that contains the lines (x,0,0) and (0,y,1)? All the planes containing (x,0,0) can be obtained by rotating the (x,y,0) plane around the X-axis. I don't think any of those will contain all of the line (0,y,1).

 

[pasmith]

They don't define one directly, but they're in the plane whose normal is given by the cross-product of line segments in each. Your condition of being in the line $z=1$ is extraneous, as it's not what I claimed.

My claim is: Given any 2 lines $L_1, L_2$ in $\mathbb R^3$, there is a plane $P$ containing $L_1, L_2$.
Proof:
Take finite segments $v_1,v_2$ along $L_1, L_2$ respectively. Then the plane $P_{12}$, whose normal is the cross product $\pm |v_1 \times v_2|$, passing through any point $p$ in either line, contains both lines $L_1, L_2$.

My example is a counterexample to this claim: The cross product of the tangents of (x,0,0) and (0,y,1) is (0,0,1), but the first does not lie in (indeed, does not intersect) z = 1 and the second does not lie in (and does not intersect) z = 0.

What is true is that any three non-colinear points define a unique plane.

 

[WWGD]

Yes, I was wrong. For one, any two lines in a plane that aren't parallel, can't be skew, so must intersect. Thus two lines that are skew but not parallel, can't be contained in a plane. Apologies, @Aurelius120 , all. Mentors, feel free to delete all my posts here.

 

[Gavran]

Mistakes are learning experiences.

 

[haruspex]

Since it is multiple choice we could cheat by trying out the offered solutions.
For the lines to be coplanar they must either intersect or be parallel (=intersect at infinity).
If k=0 is a solution then immediately z=4 and x=1, which both lead to y=2. Thus k=0 is a solution.
With k=-1, z=y+1=x+2 and x=6-z, contradicting intersection. Since the disagreement extends to the magnitude of the ratio of the x and y coefficients, we can also rule out the parallel case.
That leaves only option c.