Question

Find the vector equation of the plane determined by point ( 1,1,-1) (1, 1,1) and (0,1,1). Also find point of intersection of line r = (1+3t)I + (2-t) j + (1+t) and the plane.

Answer 1

let A be(1,1,-1)

B (1,1,1)

C (0,1,1)

then $\vec{AB} = (0,0,2)$ and $\vec{AC} = (-1,0,2)$

the vector equation will be ( Rm-Ra, AB, AC) = 0, where m (x,y,z)

coordinate form (matrix):

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

$r= \vec{Ra}+S* \vec{AB} +t* \vec{AC}$

where S and t are some coefficients

vector-coordinate form:

$\begin{cases} x= 1+(-1)*t \\ y=1 \\ z= -1 + 2*S + 2*t \end{cases}$ (1)

$line: \vec{r} = (1+3*t)* \vec{i} + (r-t)* \vec{j}+ (1+t)* \vec{k}$

$\begin{cases} x=1+3t' \\ y=2-t' \\ z=1+t' \end{cases}$ (2)

from (1) y=1, then from (2) t' = 2-y = 2-1=1

then from (2) t'= z-1 follows 1=z-1, z=2

then from (2) x=1+3*1=4

answer:

point of intersection (4,1,2)