Vector equation intersect of xy-plane

Question

Let $$L: \vec r(t)=<4-2t, -7+3t,8-13t>, \qquad t \in \mathbb{R}$$ Does this line ever intersect xy -plane? So, <4-t,-7+6t,3+2t> = < a , b, 0>, where a,b is an arbitrary constant.

Secondly,

Let $$P: \vec g(t)=<3 + t, -1 - 5t, 8t>, \qquad t \in \mathbb{R}$$ Does P ever intersect L?

Does that mean when vector r(t) - g(t) = 0??

Answer 1

Recall that the equation for the $xy$ plane is given by $z = 0$. Therefore if we have intersection of this vector parametrization and plane, there must exists a $t_0$ such that $3+2t_0 = 0$. This implies $t_0 = -3/2$ and so the intersection point is $\left(\frac{11}{2},-16,0\right)$.

To get intersection of the two lines, there must exist $s_0,t_0$ such that $g(s_0) = r(t_0)$. Now just solve the system by setting coordinates equal to each other.

Answer 2

Lines L and P do not intersect.

Let the variable in L be t1 and in P be t2. Equate the x and y coordinates. They are a system of linear equations in two variables. Solve them.

Substitute the variables in z coordinate , if they are equal then they intersect else they don't.

Here they don't.