# Thread: three questions. 3 plane

1. ## three questions. 3 plane

3.a line L in 3 space has vector equation X=(3,2,1)+ a(1,2,2). Find the distance
from P=(1,0,1) to line L
I tried use this this formula to solve d=(ax+by+cz-d)/sqrt root (a^2+b^2+c^2)
but its wrong

2，Two planes are perpendicular if they have perpendicular normal vectors. find a cartesian equations for a plane which passes through point P=(1,2,3) and is perpendicular to plane 2x+2y-z=1 and also to plane x+y-1=3.

3,The point A=(1,2,1,1),B=(2,3,0,3),C=(1,1,2,-1) in R4 from a triangle in R4. find a point D in R4 such ABCD is a parallelogram.

Thanks

2. Originally Posted by questionboy
I tried use this this formula to solve d=(ax+by+cz-d)/sqrt root (a^2+b^2+c^2)
but its wrong
That is a wrong formula. Use the following, if

$\displaystyle P=(x_0,y_0,z_0)\;,\quad Lx,y,z)=(x_1,y_1,z_1)+t(a,b,c)$

then,

$\displaystyle d(P,L)=\dfrac{\textrm{mod}\begin{vmatrix}{\vec{i}} &{\vec{j}}&{\vec{k}}\\{x_1-x_0}&{y_1-y_0}&{z_1-z_0}\\{a}&{b}&{c}\end{vmatrix}}{\sqrt{a^2+b^2+c^2} }$

Fernando Revilla

3. Originally Posted by questionboy
2，Two planes are perpendicular if they have perpendicular normal vectors. find a cartesian equations for a plane which passes through point P=(1,2,3) and is perpendicular to plane 2x+2y-z=1 and also to plane x+y-1=3.

Find a normal vector $\displaystyle \vec{v}$ to $\displaystyle (2,2,-1)$ and $\displaystyle (1,1,0)$ for example:

$\displaystyle \vec{v}=(2,2,-1)\times (1,1,0)$

Fernando Revilla

4. Originally Posted by questionboy
3,The point A=(1,2,1,1),B=(2,3,0,3),C=(1,1,2,-1) in R4 from a triangle in R4. find a point D in R4 such ABCD is a parallelogram.

Find $\displaystyle D$ such that $\displaystyle B-A=D-C$ .

Fernando Revilla