Find the cartesian equation of the plane which passes through the point
(3,-4, 1) and which is parrallel to the plane containing the point (1,2,-1) and the line r=t( i+j+k)
Hello, kingkaisai2!
Find the equation of the plane which passes through the point $\displaystyle P(3,-4,1)$ and which is
parallel to the plane containing the point $\displaystyle A(1,2,-1)$ and the line $\displaystyle \vec{r} \:=\:t(\vec{i} + \vec{j} + \vec{k})$
To write the equation of a plane, we need a point $\displaystyle (x_1,y_1,z_1)$ and its normal direction: $\displaystyle \vec{n} = \langle a,b,c\rangle$
. . Then the equation is: .$\displaystyle a(x-x_1) + b(y-y_1) + c(z-z_1)\:=\:0$
We have a point $\displaystyle P(3,-4,1)$, so we need the normal vector: $\displaystyle \vec{n}$
Since this plane is parallel to the plane containing point $\displaystyle A$ and line $\displaystyle \vec{r}$,
. . it has the same normal vector.
The line $\displaystyle \vec{r}\:=\:t(\vec{i} + \vec{j} + \vec{k})$ has parametric equations: .$\displaystyle \begin{Bmatrix}x = t \\ y = t \\ z = t\end{Bmatrix}$
For $\displaystyle t = 1$, we have the point: $\displaystyle B(1,1,1)$
For $\displaystyle t = 2$, we have the point: $\displaystyle C(2,2,2)$
Hence, the plane contains points: $\displaystyle A(3,-4,1),\;B(1,1,1),\;C(2,2,2)$
The normal vector is perpendicular to $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}.$
Since $\displaystyle \overrightarrow{AB} = \langle -2,5,0\rangle$ and $\displaystyle \overrightarrow{AC} = \langle -1,6,1\rangle$
. . $\displaystyle \vec{n}\;=\;\overrightarrow{AB} \times \overrightarrow{AC} \;= \;\begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\ -2 & 5 & 0 \\ -1 & 6 & 1\end{vmatrix} \;=\;\langle 5,2,-7\rangle$
So the desired plane has point $\displaystyle P(3,-4,1)$ and normal $\displaystyle \vec{n} = \langle 4,2,-7\rangle$
Can you finish it now?