Question

Find the vector equation of the plane determined by the points (1, 0, −1),
(0, 1, 1) and (−1, 1, 0). Also nd the point of intersection of the line

Answer 1

There are three points let's name them as

P(1,0,-1)

Q(0,1,1)

R(-1,1,0)

$\overrightarrow{PQ}=-i+j+2k \newline \overrightarrow{RQ}=i+k$

cross product of these two vector will give us the normal of plane

$\begin{vmatrix} i & j &k \\ -1 & 1&2\\1&0&1 \end{vmatrix}$$=$$i+$$3j-k$

Equation of the vector can be written as

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

$1(x-1)+3(y-0)-1(z-(-1))$

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