# Solving Vector Problem given Point & Line

I need to find the equation of the plane that contains point $(1,0,0)$ and line $r=(1+\lambda)i + 3j + 2\lambda k$

Plane parallel to $(1,3,0)-(1,0,0)=(0,3,0)$

Plane perpendicular to $(1,0,2)\times (0,3,0) = (6,0,3)$

Then

$r \cdot (2,0,1) = 2$

Try 1

So I first thought to find the normal

$n = (1,0,0) \times (1,0,-2) = (0,2,0)$

Then I have

$r\cdot (0,2,0) = (1,0,0) \cdot (0,2,0)$

$r\cdot (0,2,0) = 0$

Which is wrong?

Try 2

I also tried using the formula ${\bf r}={\bf a}+\lambda{\bf u}+\mu{\bf v}$

$r = (1,0,0) + \lambda (1,0,0) + \mu (1,0,-2)$

Which is wrong too. Why is that?

How can I express the plane in normal form and ${\bf r}={\bf a}+\lambda{\bf u}+\mu{\bf v}$? How do convert between them? And whats wrong my my tries?

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The line is $(1,3,0)+\lambda(1,0,2)$; so it contains $(1,3,0)$, and also contains $(2,3,2)$. The plane contains those two points, and also contains $(1,0,0)$. These three points are not colinear (since $(1,0,0)$ is not on the line); so the vector from $(1,0,0)$ to $(1,3,0)$; and the vector from $(2,3,2)$ to $(1,3,0)$, both lie "on" the plane. These vectors are $$\mathbf{v}=(1,3,0) - (1,0,0) = (0,3,0)$$ and $$\mathbf{w} = (2,3,2) -(1,3,0) = (1, 0, 2).$$ With these two vectors, you can get the normal to the plane by computing $$\mathbf{v}\times\mathbf{w}.$$
Your first method does not work because $(1,0,0)$ does not correspond to a vector "on" the plane, so your $n$ is not the normal of the plane ($(1,0,0)$ points from the origin to a point on the plane, but since the plane does not contain the origin, that vector does not lie "on" the plane, so it cannot help you find the normal).
Your second method does not work because $(1,0,2)$ is not a point on the plane.
I find that if I use $(0,3,0) \times (1,0,-2)$ I get $(-6,0,3)$ instead $6$ should be positive. How do I decide the order in which I compute the solution? –  Jiew Meng Mar 4 '12 at 5:28
@Jiew Meng: You computed wrong: $$(0,3,0)\times(1,0,-2) = \left|\begin{array}{ccr}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&3&0\\1&0&-2\end{array}‌​\right|=(-6-0)\mathbf{i} -(0-0)\mathbf{j} +(0-3)\mathbf{k} = (-6,0,-3).$$ In general, $\mathbf{v}\times\mathbf{w} = -(\mathbf{w}\times\mathbf{v})$. So it doesn't matter which order you do it in, since $6x+3z=0$ is the same plane as $-6x-3z=0$. –  Arturo Magidin Mar 4 '12 at 5:58