Vector-help!

**kingkaisai2** · October 28th 2006, 12:24 AM

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)

**Soroban** · October 28th 2006, 06:51 AM

Hello, kingkaisai2!

Find the equation of the plane which passes through the point $P(3,-4,1)$ and which is
parallel to the plane containing the point $A(1,2,-1)$ and the line $\vec{r} \:=\:t(\vec{i} + \vec{j} + \vec{k})$

To write the equation of a plane, we need a point $(x_1,y_1,z_1)$ and its normal direction: $\vec{n} = \langle a,b,c\rangle$
. . Then the equation is: . $a(x-x_1) + b(y-y_1) + c(z-z_1)\:=\:0$

We have a point $P(3,-4,1)$ , so we need the normal vector: $\vec{n}$

Since this plane is parallel to the plane containing point $A$ and line $\vec{r}$ ,
. . it has the same normal vector.

The line $\vec{r}\:=\:t(\vec{i} + \vec{j} + \vec{k})$ has parametric equations: . $\begin{Bmatrix}x = t \\ y = t \\ z = t\end{Bmatrix}$
For $t = 1$ , we have the point: $B(1,1,1)$
For $t = 2$ , we have the point: $C(2,2,2)$

Hence, the plane contains points: $A(3,-4,1),\;B(1,1,1),\;C(2,2,2)$

The normal vector is perpendicular to $\overrightarrow{AB}$ and $\overrightarrow{AC}.$

Since $\overrightarrow{AB} = \langle -2,5,0\rangle$ and $\overrightarrow{AC} = \langle -1,6,1\rangle$
. . $\vec{n}\;=\;\overrightarrow{AB} \times \overrightarrow{AC} \;= \;\begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\ -2 & 5 & 0 \\ -1 & 6 & 1\end{vmatrix} \;=\;\langle 5,2,-7\rangle$

So the desired plane has point $P(3,-4,1)$ and normal $\vec{n} = \langle 4,2,-7\rangle$

Can you finish it now?

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