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