How to find the equation of planes with the provided information.

Question

Two planes intersect in the line with vector equation [x,y,z] = [3,5,4] + s[2,3,1]. Point A (0,0,0) lies on the first plane and point B (1,1,1) lies on the second plane. Determine the scalar equations of the two planes?

For this type of question I would usually just do the dot product between: [2s+3, 3s+5, s+4] and B[1,1,1]. Which would let me find the value of s for which B and B' which lies on the line are perpendicular. Afterwards, I would use the cross product between BB' and [2s+3, 3s+5, s+4] to find the normal vector of the second plane. But I wouldn't know how to solve it for point A since it just gives me the value of 0=0.

Answer 1

I am having trouble following your method, but we know that if $\vec n = ( a,b,c )$ is the vector orthogonal to the plane, then the scalar equation for the plane is $ax + by + cz = d$. The most straightforward way to find the vector orthogonal to each plane is to compute the cross product of two vectors that lie on the plane. The problem gives us enough information to do this.

Starting with the first plane, we can compute vectors that point along the plane from point A to some vector that lies on the intersection of the two planes (every point on the line belongs to both planes). Lets take the vector $(3,5,4) - (0,0,0) = (3,5,4)$ and the vector $[(3,5,4) + (2,3,1)] - (0,0,0) = (5,8,5)$. $(3,5,4)$ and $(5,8,5)$ both lie on the first plane and their cross product $(3,5,4)\times (5,8,5)$ will give us the vector orthogonal to both of those vectors, i.e. $\vec n$. Computing this, you should get the vector $(-7, 5, -1)$. Now we have $-7x + 5y -z = d$j. Plugging in any point on the plane, we can solve for $d$. The point $(0,0,0)$ lies on the plane, so $d=0$ and the final equation of the first plane is $-7x + 5y -z = 0$.

I'll let you try the other plane. To recap what I did, I picked three points on the plane: point A, and two points on the line of intersection. Then I computed the vectors pointing from point A to the two other points, and used the two resulting vectors to compute the vector orthogonal to the plane via the cross product, which the scalar equation follows from.

Hope that helps and welcome to math stackexchange!

Answer 2

Let's find the equation of first plane.

Since the point $(3,5,4)$ lies on the line of intersection of the planes, it will also lie on both the planes. You also know that $(0,0,0)$ lies on the first plane. Let's assume any arbitrary point on the plane $(x,y,z)$.

The vectors $(x-0)\hat{i}+(y-0)\hat{j}+(z-0)\hat{k}$, $(3-0)\hat{i}+(5-0)\hat{j}+(4-0)\hat{k}$ and $2\hat{i}+3\hat{j}+1\hat{k}$ (the vector pointing in direction of line of intersection of planes) are co-planer. Therefore this is the equation of first plane is

$$\begin{vmatrix} x & y & z \\ 3 & 5 & 4 \\ 2 & 3 & 1 \end{vmatrix}=0$$

or

$$-7x+5y-z=0$$

The points $(3,5,4)$ and $(1,1,1)$ lies on the second plane. Assuming any arbitrary point $(x,y,z)$ on this plane.

The vectors $(x-1)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k}$, $(3-1)\hat{i}+(5-1)\hat{j}+(4-1)\hat{k}$ and $2\hat{i}+3\hat{j}+1\hat{k}$ are co-planer. Therefore this is the equation of first plane is

$$\begin{vmatrix} x-1 & y-1 & z-1 \\ 3-1 & 5-1 & 4-1 \\ 2 & 3 & 1 \end{vmatrix}=0$$

or

$$-5x+4y-2z+3=0$$

Answer 3

Your method isn’t quite right. Instead of finding a vector $B'$ on the line that’s orthogonal to $B$, you need a $B'$ such that $B'-B$ is orthogonal to the line’s direction vector. In fact, you don’t even need that much: you can use any point on the line for $B'$, then compute the cross product of $B'-B$ and the line’s direction vector to obtain a normal to the plane.

Let’s see what happens when we carry through the calculation that you’ve outlined for finding the plane through $B$: $$[2s+3,3s+5,s+4]\cdot[1,1,1] = 6s+12 = 0$$ implies that $s=-2$ and so $B'=[-1,-1,2]$ and $B'-B=[-1,-1,10]$. The cross product of this with $[2s+3,3s+5,s+4]$ is equal to $[29s+46,-19s-26,-s-2]$. This is a bit of a problem, since it still has $s$ in it. Perhaps you meant to take the cross product with the line’s direction vector instead: $[-1,-1,10]\times[2,3,1] = [29,-19,-1]$. The point-normal Cartesian equation of the plane is then $29x-19y-z=[29,-19,-1]\cdot[1,1,1]=9$. Since we used $B$ to construct this equation, we know that this point satisfies it, but every point on the line must also lie on this plane. Unfortunately, $[29,-19,-1]\cdot[2s+3,3s+5,s+4]=-12\ne9$, so no point of the line lies on this plane.

Instead, let’s take $B'=[3,5,4]$ and compute $(B'-B)\times[2,3,1]=[-5,4,-2]$, yielding the equation $5x-4y+2=[5,-4,2]\cdot[1,1,1]=3$ for the plane. This time, $[5,-4,2]\cdot[2s+3,3s+5,s+4]=3$, so every point on the line satisfies this equation, as required.

Using this method, there should be no trouble with zeros any more. Taking again $A'=[3,5,4]$, a normal to the plane through $A$ is $[3,5,4]\times[2,3,1]=[-7,5,-1]$ and since this plane passes through the origin, an equation for it is simply $7x-5y+z=0$.

---

There are a few alternative methods that allow you to compute equations for the two planes directly, without having to solve for any unknowns. The vectors $[3,-2,0]$ and $[0,1,-3]$ are both perpendicular to the line’s direction, so, using the point-normal form of equation with the point $[3,5,4]$ on the line, two planes that include this line are given by the equations $3x-2y+1=0$ and $y-3z+7=0$. Every plane that includes this line has an equation that’s an affine combination of these two equations: $$(1-\lambda)(3x-2y+1)+\lambda(y-3z+7)=0.$$ To find the equation of the plane that passes through a given point not on the line, we can substitute its coordinates and solve for $\lambda$, or compute it directly using a technique called Plücker’s mu: If the point has coordinates $(x_0,y_0,z_0)$, the the equation of the plane that passes through this point and the line is $$(y_0-3z_0+7)(3x-2y+1)-(y-3z+7)(3x_0-2y_0+1)=0.$$ For example, this formula gives for the plane through $A$ $$7(3x-2y+1)-(y-3z+7) = 21x-15y+3z=0,$$ which is clearly equivalent to the equation derived earlier.

You could also write the equations down directly in the form of determinants, as explained in this answer. It’s even simpler (in some ways) if you work in homogeneous coordinates to do this. Append a $1$ to the coordinates of each finite point on the plane and a $0$ to the line’s direction vector to form its point at infinity, which also lies on the plane, then form a $4\times4$ determinant. Working again with the point $A$, this produces the equation $$\begin{vmatrix}x&y&z&1\\3&5&4&1\\2&3&1&0\\0&0&0&1\end{vmatrix} = -7x+5y-z=0,$$ which is again obviously equivalent to the equations for this plane found previously. Equivalently, delete the first row of this matrix and compute an element $\mathbf n$ of the null space of the remaining $3\times4$ matrix. An equation of the plane is then $\mathbf n\cdot[x,y,z,1]=0$.