# Find the point at which the given lines intersect. Find an equation of the plane that contains these lines.

-The two lines with the following equations intersect at a point. $r=(2,3,0) + t (3,-3,2)$ $r=(5,0,2) + s (-3,3,0)$ -(a) Find out the point of intersection of these two lines. -(b) Find the equation of the plane having these two lines. In this question, we have to find two things: the point of intersection and the equation of the plane in which the given two lines lie. The basic concept behind this article is the knowledge of vectors, normal vectors, the cross-product of two vectors, and a sound understanding of the equation of a plane.

Part (a) – To find out the point of intersection, we will put both the equations of vectors equal to each other so that we can get values of unknown variables $t$ and $s$. Given vectors are as follows: $r= (2,3,0) + t (3,-3,2)$ $r= (5,0,2) + s (-3,3,0)$ Vector $1$ will be: $r= (2,3,0) + t (3,-3,2) = (2 + 3t, 3 – 3t, 2t )$ Vector $2$ will be: $r= (5,0,2) + s (-3,3,0) = (5 – 3s, 3s, 2)$ Putting two vectors equal, by putting the first equation equal to the first of the other vector: $2 + 3t = 5 -3s$ Putting two vectors equal, by putting the second equation equal to the second of the other vector: $3 – 3t = 3s$ Putting two vectors equal, by putting the third equation equal to the third of the other vector: $2t = 2$ From the above equation, we get the value of $t$, which is: $2t = 2$ $t= \dfrac { 2}{ 2}$ $t =1$ Now putting the value of $t$ in any of the above equations, we get the value of $s$. Putting it in the following equation: $3-3t=3s$ $3 – 3(1) = 3s$ $3 – 3 = 3s$ $s = 0$ We can also verify the value of $s=0$ by putting the value of $t$ in the other equation as follows: $2+3t=5-3s$ Put $t=1$ $2+3t=5-3s$ $2+3(1)=5-3s$ $5=5-3s$ $-3s=5-5$ $s=0$ Now putting the values of $t$ and $s$ in the coordinates, we will get the point of intersection: $(2 + 3(1),3-3(1),2(1)) \space; (5-3(0),3(0),2)$ $(5,0,2) \space; (5,0,2)$ Part (b) – To find the equation of the plane, we should know the normal vector so taking the cross product: $=<3,-3,2> \times <-3,3,0>$ $=<-9,-9,0>$ The non-zero scalar multiple of this vector is also a normal vector, so: $=<-1,-1,0>$ We know that the normal vector is perpendicular to directional vectors of given lines in the plane; we can write the equation with the point of intersection as: $1(x-5) +1(y-0)+ 0(z-2)=0$ $x-5+y=0$

## Numerical Results

Part (a) – Point of intersection is: $(5,0,2)$ Part (b) – Equation of plane is: $x-5+y=0$

## Example

Find out the point of intersection of these two lines: $r=(7,3,0)+t(0,-1,2)$ $r=(4,0,2)+s(3,2,0)$ $r=(7+0t,3-t, 2t)$ $r=(4+3s,2s, 2)$ Putting two vectors equal: $7=4+3s$ $3-t=2s$ $2t=2$ $t=\dfrac{2}{2}=1$ $3-(1)=2s$ $s=1$ Point of intersection: $(7+ 0(1),3-(1),2(1)) \space; (4+3(1),2(1),2)$ $(7,2,2) \space; (7,2,2)$