If we are given two lines 
Coplaner then how can we find the value of k.
I think if they are coplaner then their cross product should be zero .
In the solution it is given as 
But I could not understand what they have done.
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what means $$\frac{y-2}{0}$$? – Dr. Sonnhard Graubner Dec 25 '16 at 12:25
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Any two vectors are coplanar with each other (since a set of two vectors defines a plane). So are you saying that the cross product of any two vectors is zero? – probably_someone Dec 25 '16 at 12:26
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Your question's title says "intersection", your question says coplanar, you have $;\frac{y-2}0;$ ...check your data and say clearly what you want!! – DonAntonio Dec 25 '16 at 12:26
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@Dr.SonnhardGraubner: $\frac{y-2}{0}$ means that the second component of the directional vector is zero. It's the symmetric form of the equation of lines. In the United States it's common to rewrite symmetric equation without this division by zero as $\frac{x-1}{2}=\frac{z+1}{k}, y=2$ -- with a separate equation for the variable whose value is constant. But in some other countries mathematical tradition allows writing the form given in the OP's question. It's understood kinda formally as simply providing the coefficients of the directional vector in a certain layout. – zipirovich Dec 25 '16 at 17:48
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Hint: Examine both lines in parametric form. If their vectors are parallel then they are certainly coplanar. If their vectors are not parallel, two lines are coplanar if and only iff they intersect; otherwise, they are skew. Hope it helps.
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Did you pay attention to the fact your hint is for a question whose data doesn't make sense? – DonAntonio Dec 25 '16 at 12:29
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@DonAntonio I know that but still I want him to know the condition for two lines to be coplanar. Then I thought maybe he can apply it to any question of this type. – Dec 25 '16 at 12:30
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@DonAntonio: Sorry, but -1 to you as well as to Rohan. See the comment above. The given data makes perfect sense. – zipirovich Dec 25 '16 at 17:49
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@zipirovich Sorry but I didn't post an answer here so you can't downvote me...and no: dividing by zero makes no sense at all, though I understand you can believe it does. If the OP had something to say about this he could have replied long time ago. – DonAntonio Dec 25 '16 at 18:49
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@DonAntonio: Please pay attention. I didn't downvote you. I couldn't have possibly done that, as this answer wasn't even yours, and there's no such thing as downvoting on comments. I wrote "-1" rather as a figure of speech... Respectfully, – zipirovich Dec 25 '16 at 18:59
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And on the topic of representing equations -- if you're unaware of something, that doesn't mean that it doesn't exist. The whole point of my comment was to point out that there's a tradition of mathematical notation, well and alive, that allows this form of writing equations even though when taken literally they don't make sense. – zipirovich Dec 25 '16 at 19:05
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@zipirovich I won't insist on this as I honestly think it isn't worth our, both yours and mine, time. Anyway, would you mind naming some of those countries you write about in your comment below the question who use the notation "division by zero" to denote some component equals zero? Some few links to books or papers on this will help. Thanks. – DonAntonio Dec 25 '16 at 21:55
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@zipirovich Still no "some countries" which use that? And the OP still hasn't said half a thing.. – DonAntonio Dec 26 '16 at 06:09
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I think if they are coplaner then their cross product should be zero.
This statement is at least unclear, and not quite meaningful or correct. First of all, what do "they" and "their" refer to? If you mean the lines, then this statement is meaningless because cross product is an operation on vectors, not lines. It would be meaningful when talking about vectors, but then it's not true — the cross product of two (nonzero) vectors is zero means that they are collinear, not coplanar.
Let $\vec{a}$ and $\vec{b}$ be the directional vectors of two lines, and $A$ and $B$ some points on the same lines, respectively. (All of these can be seen from the given equations.) The condition given in the answers requires that the three vectors $\vec{a}$, $\vec{b}$, and $\vec{AB}$ be coplanar to guarantee that the two lines are. To check whether they are coplanar, we can set up the triple scalar product of these vectors. Effectively it finds the volume of the parallelepiped (up to the sign) built on these three vectors as sides. If the volume is zero, the parallelepiped is "flat" due to the three vectors lying in the same plane.
Why do we want $\vec{a}$, $\vec{b}$, and $\vec{AB}$ to be coplanar? Consider two cases. (1) If $\vec{a}\parallel\vec{b}$, then the lines are parallel and therefore coplanar. Note that in this case the three vectors are also coplanar regardless of the third vector. (2) Otherwise, we need to distinguish between intersecting lines (coplanar) and skew lines (not coplanar). If the lines are intersecting, then all their points lie in the same plane as $\vec{a}$ and $\vec{b}$, therefore $\vec{AB}$ must lie in that same plane.
zipirovich
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