Answer to Question #197943 in Analytic Geometry for Dhruv rawat

Question

Find the vector equation of the plane determined by the points (0,2,1),(2,1,0),(1,-1,0). Further, check whether the line r=(1+2α)I + (2-3α)j - (1+5α)k intersects this plane. If it intersects, find the point of intersection. If the line and plane do not intersects,find the equation of another line that intersects this plane.

Answer 1

$A(0,2,1), B(2,1,0),C(1,-1,0)\\ \begin{vmatrix} x-0 & y-2&z-1 \\ 2-0 & 1-2&0-1\\ 1-0&-1-2&0-1 \end{vmatrix}=0\\$

$\begin{vmatrix} x & y-2&z-1 \\ 2 & -1&-1\\ 1&-3&-1 \end{vmatrix}=0\\$

$-2x+y-5z+3=0$

The vector equation of the plane:

$(0,2,1)+v(2-0,1-2,0-1)+\\ +u(1-0,-1-2,0-1)=\\ =(2v+u,2-v-3u,1-v-u)$

$r=(1+2a)i+(2-3a)j-(1+5a)k=\\ =(1+2a,2-3a,-(1+5a))\\ -2(1+2a)+(2-3a)-5(-1-5a)+3=0\\ 18a+8=0\\ a=-\frac{4}{9}\\ x=1+2\cdot\frac{-4}{9}=\frac{1}{9}\\ y=2-3\cdot \frac{-4}{9}=\frac{10}{3}\\ z=-(1+5\cdot \frac{-4}{9})=\frac{11}{9}\\ (\frac{1}{9}, \frac{10}{3},\frac{11}{9})$