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**I-Think** GCE Maths

With respect to an origin O, the pts. A and B have the following position vectors

OA=2i+2j+k OB=i+4j+3k

The line has vector equation

r=4i-2j+2k+s(i+2j+k)

I. Prove that the line l does not intersect the line through A and B

II. Find the equation of the plane containing l and the point A giving your answer in the form ax+by+cz=d

First we need to find the vector equation of the line through $\displaystyle A$ and $\displaystyle B$. This can be written as$\displaystyle \vec r = \vec{OA}+t\vec{AB}$ ...(1)

where $\displaystyle t$ is a scalar parameter. Do you understand this bit? It simply says:$\displaystyle \underbrace{\vec r=}_{\text{You can move from O to any point on the line ... }}$$\displaystyle \underbrace{\vec{OA}}_{\text{ by a movement to A ...}}\underbrace{+}_{\text{ followed by ...}}\underbrace{t\vec {AB}}_{\text{ a movement in the direction of AB}}$

Now$\displaystyle \vec{OA} = 2\vec i + 2\vec j + \vec k$

and$\displaystyle \vec{AB} = \vec{OB} - \vec{OA}$$\displaystyle = -\vec i +2 \vec j +2\vec k$

So, from (1), the equation of the line through $\displaystyle AB$ is:$\displaystyle \vec r = 2\vec i + 2\vec j + \vec k+t(-\vec i +2 \vec j +2\vec k)$

Next, we need to prove that this line doesn't intersect line $\displaystyle l$. We do this by showing that there are no values of $\displaystyle s$ and $\displaystyle t$ that will bring us to the same point on each of the lines. So, suppose that the lines meet. Then, for some value of $\displaystyle s$ and of $\displaystyle t$:$\displaystyle 2\vec i + 2\vec j + \vec k+t(-\vec i +2 \vec j +2\vec k)= 4\vec i-2\vec j+2\vec k+s(\vec i+2\vec j+\vec k)$

Comparing the coefficients of $\displaystyle \vec i$ and $\displaystyle \vec j$, we get:$\displaystyle 2-t = 4+s$

and$\displaystyle 2+2t = -2+2s$

which gives:$\displaystyle s = 0,\;t=-2$

If we now compare the coefficients of $\displaystyle \vec k$:On the LHS: $\displaystyle 1-4 = -3$

On the RHS: $\displaystyle 2 + 0 = 2$

Since these are not equal, there are no values of $\displaystyle s$ and $\displaystyle t$ that satisfy the equation, and hence the lines do not intersect.

Part II

We have a choice of two methods here. The first uses simple algebra; the second, rather more sophisticated, uses a vector and scalar product.

For the first method, we find a couple of points on the line $\displaystyle l$ by taking two values of $\displaystyle s$, and then solve a set of simultaneous equations.

So, with $\displaystyle s = 0$ and $\displaystyle s = 1$, we get the points $\displaystyle (4, -2, 2)$ and $\displaystyle (5, 0, 3)$ lying on $\displaystyle l$, and therefore also in the plane. And the point $\displaystyle A$ also lies in the plane.

So, if the plane is $\displaystyle ax+by+cz=1 $ (which is simply the equation $\displaystyle ax+by+cz=d$ re-written when divided throughout by $\displaystyle d$, using new values of $\displaystyle a, b, c$), we get:$\displaystyle \left \{\begin{array}{l}

4a-2b+2c=1\\

5a+0b+3c=1\\

2a+2b+c=1\\

\end{array}\right.$

This gives:$\displaystyle \left \{\begin{array}{l}

a=1\\

b=\tfrac16\\

c=-\tfrac43\\

\end{array}\right.$

and, when we get rid of fractions, gives the plane:

$\displaystyle 6x+y-8z=6$

If you want to see an explanation of the vector method, look at this page.

Grandad