Question #137169
What is the relationship between the vector (2,1,2) and the line. L defined by the following parametric equations
x=1+2t
y=2+t
z=3-2t
1
Expert's answer
2020-10-08T12:34:07-0400

The general vector equation of a straight line is,(x,y,z)=a+λdwhereλis a parameter anddis the direction vector.Let the vector equation of the line be defined asL1.(x,y,z)=(1,2,3)+(2,1,2)tThe vector equation of the line perpendicular toL1passing through lineP(x0,y0,z0)is given byax+by+cz=ax0+by0+cz0.....(1)Define the vector equation of the lineperpendicular to line passing throughPasL2.(a,b,c)in(1)is the direction vectorof the line passing throughL1,thus,(a,b,c)=(2,1,2).2x+y2z=ax0+by0+cz0.IfL2passes throughM(2,1,2),2x+y2z=(2,1,2)(2,1,2)=4+14=1=1Thus,2x+y2z=1passes through the point(2,1,2)and is perpendicular toL1.To find the point where the linesx(t),y(t)andz(t)intersects the plane,substitute the parametric equationsof the vector equation of the line2x+y2z=12(1+2t)+(2+t)2(32t)=12+4t+2+t6+4t=12+9t=19t=3,t=13Thus, att=13,L1intersectsL2The coordinates of a point N onL1att=13aregiven by(53,73,73).Thus, the vector equation of the line passingthrough pointM&Nis the line passing through(2,1,2)and(53,73,73)which is given by(x,y,z)=(2,1,2)+λ((53,73,73)(2,1,2))(x,y,z)=(2,1,2)+λ(13,43,13)......(2)Thus, equation(2)shows the relationshipbetween(2,1,2)and the vector equation of the lineL1.\textsf{The general vector equation of a straight line is,}\\ (x, y, z) = a + \lambda d \hspace{0.1cm}\textsf{where} \hspace{0.1cm} \lambda\hspace{0.1cm}\textsf{is a parameter and}\\d\hspace{0.1cm} \textsf{is the direction vector.}\\ \textsf{Let the vector equation of the line be defined as}\hspace{0.1cm} L_1.\\ (x, y, z) = (1, 2, 3) + (2, 1, -2)t\\ \textsf{The vector equation of the line perpendicular to}\hspace{0.1cm} L_1\\\textsf{passing through line}\hspace{0.1cm} P(x_0, y_0, z_0)\hspace{0.1cm}\textsf{is given by}\\ ax + by + cz = ax_0 + by_0 + cz_0\hspace{0.1cm}..... (1)\\ \textsf{Define the vector equation of the line}\\\textsf{perpendicular to line passing through}\hspace{0.1cm}P \hspace{0.1cm}\textsf{as}\hspace{0.1cm} L_2. \\ (a, b, c)\hspace{0.1cm}\textsf{in}\hspace{0.1cm}(1) \hspace{0.1cm}\textsf{is the direction vector}\\\textsf{of the line passing through}\hspace{0.1cm}L_1, \hspace{0.1cm}\textsf{thus,}\hspace{0.1cm} (a, b, c) = (2, 1, -2).\\ \therefore 2x + y - 2z = ax_0 + by_0 + cz_0.\\ \textsf{If} \hspace{0.1cm}L_2 \hspace{0.1cm}\textsf{passes through}\hspace{0.1cm} M(2, 1, 2),\\ \begin{aligned} 2x + y - 2z &= (2, 1, -2) \cdot (2, 1, 2)\\&= 4 + 1 - 4 = 1\\&=1 \end{aligned}\\ \textsf{Thus,}\hspace{0.1cm} 2x + y - 2z = 1\hspace{0.1cm} \textsf{passes through the point}\hspace{0.1cm}(2, 1, 2) \\\textsf{and is perpendicular to}\hspace{0.1cm} L_1.\\ \textsf{To find the point where the lines}\hspace{0.1cm} x(t), \\y(t) \hspace{0.1cm}\textsf{and}\hspace{0.1cm} z(t)\hspace{0.1cm} \textsf{intersects the plane,}\\\textsf{substitute the parametric equations}\\\textsf{of the vector equation of the line}\\ 2x + y - 2z = 1\\ 2(1 + 2t) + (2 + t) - 2(3 - 2t) = 1\\ 2 + 4t + 2 + t - 6 +4t = 1\\ -2 + 9t = 1\\ \displaystyle 9t = 3, t = \frac{1}{3}\\ \displaystyle\textsf{Thus, at}\hspace{0.1cm} t = \frac{1}{3}, \hspace{0.1cm} L_1 \hspace{0.1cm}\textsf{intersects}\hspace{0.1cm}L_2 \\ \displaystyle\textsf{The coordinates of a point N on}\hspace{0.1cm} L_1 \hspace{0.1cm} \textsf{at}\hspace{0.1cm}t = \frac{1}{3} \hspace{0.1cm}\textsf{are}\\\textsf{given by}\hspace{0.1cm}\left(\frac{5}{3}, \frac{7}{3},\frac{7}{3}\right).\\ \displaystyle\textsf{Thus, the vector equation of the line passing}\\\textsf{through point}\hspace{0.1cm} M\hspace{0.1cm}\&\hspace{0.1cm} N \\\textsf{is the line passing through}\hspace{0.1cm}(2, 1, 2)\hspace{0.1cm}\textsf{and}\\\left(\frac{5}{3}, \frac{7}{3}, \frac{7}{3}\right) \hspace{0.1cm}\textsf{which is given by}\\ \displaystyle(x, y, z) = (2, 1, 2) + \lambda\left(\left(\frac{5}{3}, \frac{7}{3}, \frac{7}{3}\right) - (2, 1, 2)\right)\\ \displaystyle(x, y, z) = (2, 1, 2) + \lambda\left(\frac{-1}{3}, \frac{4}{3}, \frac{1}{3}\right) \hspace{0.1cm}......(2)\\ \textsf{Thus, equation}\hspace{0.1cm} (2) \hspace{0.1cm}\textsf{shows the relationship}\\\textsf{between} \hspace{0.1cm}(2, 1, 2) \hspace{0.1cm}\textsf{and the vector equation of the line}\hspace{0.1cm}L_1.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS