# Math Help - Vector intersection

1. ## Vector intersection

I was asked to find the normal vector and the Cartesian equation of the plane that contains the three points A(1, 1, 1,), B(2, 3, 4) and C(1, 2, 2). I've found that the normal vector is $\mathbf{i}-\mathbf{j}+\mathbf{k}$ and that the Cartesian equation is $x-y+z = 3$. Now I'm asked whether the line r = t[1 0 0] intersects this plane. How do you find that?

2. Originally Posted by Penchant
I was asked to find the normal vector and the Cartesian equation of the plane that contains the three points A(1, 1, 1,), B(2, 3, 4) and C(1, 2, 2). I've found that the normal vector is $\mathbf{i}-\mathbf{j}+\mathbf{k}$ and that the Cartesian equation is $x-y+z = 3$. Now I'm asked whether the line r = t[1 0 0] intersects this plane. How do you find that?
Dear Penchant,

First of all I think you have done some mistakes in your calculations.

$A=(1,1,1)$

$B=(2,3,4)$

$C=(1,2,2)$

$\overline{AB}\times\overline{AC} = -\underline{i}-\underline{j}+\underline{k}$

Therefore the normal vector is $-\underline{i}-\underline{j}+\underline{k}$

Using this result you would get the equation of the plane as, $x+y-z=1$

Since, $r=t\left[\begin{array}{cc}1\\0\\0\end{array}\right]$

$x=t , y=0 , z=0$

Therefore the line coincides with the x axis.

Since (1,0,0) is on the plane which we are considering the line intersects the plane.

3. Thank you, Sudharaka. You are right. I made a mistake.
Using this result you would get the equation of the plane as, $x+y-z=1$
$r \cdot\left[\begin{array}{cc}-1\\-1\\1\end{array}\right] = \left[\begin{array}{cc}2\\3\\4\end{array}\right]\left[\begin{array}{cc}-1\\-1\\1\end{array}\right] = -2-3+4$
$\Rightarrow r \cdot\left[\begin{array}{cc}-1\\-1\\1\end{array}\right]$

$\Rightarrow -x-y+z = -1 \Leftrightarrow x+y-z=1.$

Right.

But I think I need a little bit more explanation on this bit:
Since, $r=t\left[\begin{array}{cc}1\\0\\0\end{array}\right]
$

$

x=t , y=0 , z=0
$
Where does the intersection happen at?

4. Dear Penchant,

The intersection happens at (1,0,0). That is when t=1.

5. Dear Penchant,

Did you fully understand everything now?

6. Originally Posted by Sudharaka
Dear Penchant,

Did you fully understand everything now?
You are a wonderful person.

I just have little problem figuring out how you see the intersections. Can we find the point of intersection through a direct calculation?

I ask this because I'm as well having problem with figuring whether the line M that passes through the points A(2, -1, 4) and B(2, 5, 6) intersects the plane with the vector form equation r = [1 0 3]+t[2 3 1]. I've found that vector equation of M is r = [1, 0, 3]+t[2, 3, 1], but again have problem with finding the intersections.

7. Dear Penchant,

The equation r = [1 0 3]+t[2 3 1] is a line equation in vector form. It is not a plane. For a plane the general vector equation is $(\underline{r}-\underline{a}).\hat{\underline{n}}=0$ where $\hat{n}$ is a normal vector to the plane.

Also you have figured out the equation of M incorrectly.

$\underline{r}-(2,-1,4)=t(2, 5, 6)\Rightarrow{\underline{r}=(2,-1,4)+t(2, 5, 6)}$ ; where t is a real parameter.

Now if you want to find the intersection of these two lines, the argument is that the position vector at the intersection point must be equal.

Therefore, $(1,0,3)+t_{1}(2,3,1)=(2,-1,4)+t_{2}(2,5,6)$

Then you would get three equations with two unknowns,

$1+2t_{1}=2+2t_{2}$---------1

$3t_{1}=-1+5t_{2}$----------2

$3+t_{1}=4+6t_{2}$----------3

Solving these equations you would get $t_{1}=-\frac{1}{2}$ and $t_{2}=-\frac{1}{10}$

In this case the values for $t_{1}~and~t_{2}$ satisfies both equations, therefore the two lines intersect at the point $\underline{r}=(1,0,3)+\frac{-1}{2}(2,3,1)=(0,-3/2,5/2)$

But what if the values obtained for $t_{1}~and~t_{2}$ does not satisfy the three equations.(they will satisfy only two equations.) Then we could conclude that the two lines does not intersect.