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Determine whether the points $P$ and $Q$ lie on the given surface.
$\mathbf{r}(u, v)=\langle u+v, u-2 v, 3+u-v\rangle \quad P(4,-5,1), Q(0,4,6)$

$r\left(u, v\right)=\left\langle u+v, u-2 v, 3+u-v\right\rangle$

$P(4,-5,1)$
$Q(0,4,6)$

(1) $P(4,-5,1)$

$u+v=4 \quad, \quad u-2 v=-5 \quad, \quad 3+u-v=1$

$3 v=9 \quad v=3, u=1$

check: $3+u-v=3+1-3=1$ #

$p$ on the surtice

(2) $Q(0,4,6)$

$u+v=0 \quad , \quad u - 2 v=4 \quad , \quad 3+u-v=6$

$u=\frac{4}{3} \quad v=\frac{-4}{3}$

$3+\frac{4}{3}-\left(-\frac{4}{3}\right) \neq 6$

$Q$ doesn't lie on surface

Identify the surface with the given vector equation.
$\mathbf{r}(u, v)=(u+v) \mathbf{i}+(3-v) \mathbf{j}+(1+4 u+5 v) \mathbf{k}$

$r(u, v)=(u+v) i+(3-v) j+(1+4 u+5 v) k$

$a_{0}=\left(0 , 3 , 1\right)$

$a=(1,0,4)$

$b=(1,-1,5)$

$a \times b=\left|\begin{array}{ccc}{i} & {j} & {k} \\ {1} & {0} & {4} \\ {1} & {-1} & {5}\end{array}\right|=(0+4) i+(5-4) j+(-1) k$

$=4 i-j-k$

equation of plan

$4[x-0]-1[y-3]-1[z-1]=0$

$4 x-y-z=-4$