1.a) A plane $S_{1}$ contains the three points $(1,0,0),(0,1,0)$ and $(0,0,1) .$ Find an equation for $S_{1}$

Question

1.a) A plane $S_{1}$ contains the three points $(1,0,0),(0,1,0)$ and $(0,0,1) .$ Find an equation for $S_{1}$

1b) The perpendicular from the origin $\mathrm{O}$ to another plane, $\mathrm{S}_{2}$ meets it at $(1,1,0) .$ Find an equation for $\mathrm{S}_{2}$

My Answer

1.a)

Let plane be $a x+b y+c z=d$

$\left| \begin{array}{ccc}{x-x_{A}} & {y-y_{A}} & {z-z_{A}} \\ {x_{B}-x_{A}} & {y_{B}-y_{A}} & {z_{B}-z_{A}} \\ {x_{C}-x_{A}} & {y_{c}-y_{A}} & {z_{C}-z_{A}}\end{array}\right|=0$

$\left| \begin{array}{ccc}{x-1} & {y-0} & {z-0} \\ {0-1} & {1-0} & {0-0} \\ {0-1} & {0-0} & {1-0}\end{array}\right|=0$

$\left| \begin{array}{ccc}{x-1} & {y-0} & {z-0} \\ {-1} & {1} & {0} \\ {-1} & {0} & {1}\end{array}\right|=0$

$(x-1)(1 \cdot 1-0 \cdot 0)-(y-0)((-1) \cdot 1-0 \cdot(-1))+(z-0)((-1) \cdot 0-1 \cdot(-1))=0$

$1(x-1)+1(y-0)+1(z-0)=0$

$x+y+z-1=0$

1.b)

normal vector and coordinates of point lying on plane

I have my point A (1,1,0)

but how do I find $\overline{n}$ the normal to work it out

Answer 1

Since you have 3 coordinate points on a plane, you can calculate two direction vectors that lie on the plane.

Say $x_1=(1,0,0), x_2=(0,1,0), x_3=(0,0,1)$, then we want to calculate the direction vectors from one point to the other two. i.e. arbitrarily choose a point $x_1$ and we want to calculate the direction of the vector from $x_1$ to $x_2$ and also from $x_1$ to $x_3$.

$$x_{12} = x_2 - x_1 = (-1, 1, 0), \quad x_{13} = x_3 - x_1 = (-1, 0, 1).$$

Now we can calculate the direction of the normal to the plane by taking the cross product of $x_{12}$ and $x_{13}$.

$$(x_{12}) \times (x_{13})= (1,1,1).$$

Now we have our coefficients $a,b,c$. All that is left to do now is to calculate $d$. To do this, we take one of the points that lies on the plane and substitute it into our equation for the plane then solve for $d$.

Our plane equation is $x+y+z=d$. Now arbitrarily take $x_1$ and substitute,

$$1+0+0=d=1.$$

So $d=1$ and this gives us the equation of the plane:

$$x+y+z=1.$$

Now for part b.

We have some vector from $O$ which has the direction of the normal to plane $S_2, (1,1,1)$. We can express this as the position vector, $r(t)$, where $t$ is some parameter,

$$r(t) = (1,1,0)+t(1,1,1).$$

Then we can form,

$$(\bar{a} , \bar{b} , \bar{c})- t(1,1,1) = (1,1,0).$$

Solving this gives us the plane equation for $S_2$ of

$$\bar{a} x + \bar{b} y + \bar{c} z = (1+t)x+(1+t)y+tz = \bar{d}.$$

Take $t=1$, then we can solve for $\bar{d}$ like in part a.

$$2*(1)+2*(1)+1*(0)=\bar{d}=4$$

We finally have an equation for the plane $S_2$,

$$2x+2y+z=4.$$