
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: \[(7+ 0(1),3-(1),2(1)) \space; (4+3(1),2(1),2)\] \[(7,2,2) \space; (7,2,2)\]Previous Question < > Next Question
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