-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.
Expert Answer
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:\[(7+ 0(1),3-(1),2(1)) \space; (4+3(1),2(1),2)\]\[(7,2,2) \space; (7,2,2)\]
