# Find the normal vectors of the two angle-bisecting planes.

Compute the coordinate equation of the angle bisectors of the planes E and F.

$$E: x + 4y + 8z + 50 = 0$$ and $$F: 3x + 4y + 12z + 82 = 0$$

Proceed as follows:

a) Find the normal vectors of the two angle-bisecting planes.

b) Find a shared point of planes E and F.

c) Now determine the equations of the two angle-bisecting planes.

I have the solutions but I don't understand why I must do things the way the solution is shown.

a) $$\left|\begin{pmatrix}1\\ 4\\ 8\end{pmatrix}\right|=9$$ and $$\left|\begin{pmatrix}3\\ 4\\ 12\end{pmatrix}\right|=13$$

are the normal vectors from the equations. But this is not a good enough answer, all they asked for is the normal vectors, aren't these the normal vectors? Why must I add and subtract them like this?:

$$13\cdot \begin{pmatrix}1\\ 4\\ 8\end{pmatrix}+9\cdot \begin{pmatrix}3\\ 4\\ 12\end{pmatrix}$$

$$13\cdot \begin{pmatrix}1\\ 4\\ 8\end{pmatrix}-9\cdot \begin{pmatrix}3\\ 4\\ 12\end{pmatrix}$$

b) The solution says that I must choose one component, e.g: $$x=2$$ and then I substitute it into the equations and complete the simultaneous equation to find the point. Must it be only the x component? And why the value 2? Can it be any value? So according to the solution, the shared point is $$P(2, 5, -9)$$

c) The solution uses the answers from part a and b and gets this $$\begin{pmatrix}10\\ 22\\ 53\end{pmatrix}\cdot \left[\begin{pmatrix}x\\ y\\ z\end{pmatrix}-\begin{pmatrix}2\\ 5\\ -9\end{pmatrix}\right]$$ $$\begin{pmatrix}-7\\ 8\\ -2\end{pmatrix}\cdot \left[\begin{pmatrix}x\\ y\\ z\end{pmatrix}-\begin{pmatrix}2\\ 5\\ -9\end{pmatrix}\right]$$

Is it a general rule to use the normal to find the equation from a shared point?

• Part (a) asks for the normals of the bisectors, not of the original two planes. Besides, the two expressions that you’ve written down are the norms of those normal vectors, which wasn’t at all what was required.
– amd
Dec 24, 2019 at 6:04
• The formula where they get the normals of the bisectors, is it standard? It's something I must memorize?@amd Dec 24, 2019 at 7:03
• Yes, it’s pretty standard, but you can easily reconstruct it once you understand what’s going on.
– amd
Dec 24, 2019 at 8:20

(a)The hidden process going on is to convert the direction ratios of normal to direction cosines and then take their sum, when you do it, the first plane becomes, $$\frac{x+4y+8z+50}{9}=0$$ $$\frac{3x+4y+12z+82}{13}=0$$ Now, taking the direction ratio of the angle bisectors are the sum and difference of $$\pmatrix{\frac{1}{9}\\\frac{4}{9}\\\frac{8}{9}}$$ and $$\pmatrix{\frac{3}{13}\\\frac{4}{13}\\\frac{12}{13}}$$. These are $$\frac{1}{9\times 13}$$ times the direction ratios you obtain.

(b)Nothing is compulsory. Choose any value of any coordinate you want and solve for other two.

(c)No, is not general rule, but it is the easiest method. You can use any other method you know.

For part (a) you were supposed to come up with normal vectors for the angle bisectors, not for the original two planes. Moreover, the expressions that you’ve got there are the norms of the vectors, which wasn’t at all what was being asked for.

The normals of the angle bisectors of the two planes are themselves the angle bisectors of the planes’ normals. If you have a pair of vectors $$\mathbf v$$ and $$\mathbf w$$ with $$\lVert\mathbf v\rVert=\lVert\mathbf w\rVert$$, then finding these bisectors is easy: they’re just $$\mathbf v\pm\mathbf w$$. You can verify this by drawing the usual paralellogram diagram for the sum of two vectors. When the two vectors are of equal length, this paralellogram is a rhombus and its diagonals are vertex angle bisectors. When the two vectors are of different lengths, we can adjust their lengths to be the same, then take the sum and product of the adjusted lengths. Normalizing the two vectors is an obvious way to do this, obtaining $${\mathbf v\over\lVert\mathbf v\rVert}\pm{\mathbf w\over\lVert\mathbf w\rVert},$$ or, after multiplying by the two norms, $$\lVert\mathbf w\rVert\mathbf v\pm\lVert\mathbf v\rVert\mathbf w.$$ These are exactly the expressions that you asked about.

For part (b), you can of course choose any of the variables and set it to any convenient value to reduce the number of variables in the system. Zero is a popular choice. If I had to guess, $$x=2$$ was chosen to make all of the coefficients in the resulting equations even.

Presumably, you’re meant to use the point-normal form of plane equation in part (c) because that’s what was covered in the preceding material in whatever text you’re using. There are certainly other ways to derive an equation of a plane. For instance, if you know three noncolinear points on the plane, an equation for it can be written down directly in the form of a certain determinant.

This certainly isn’t the only way to find the angle bisectors, nor is it even the easiest. In fact, equations for them can be written down directly from the equations of the two planes by using the standard point-plane distance formula: the two angle bisectors are the locus of points that are equidistant from the two planes. This gives us $${\lvert x+4y+8z+50\rvert \over \sqrt{1^2+4^2+8^2}} = {\lvert 3x+4y+12z+82\rvert \over \sqrt{3^2+4^2+12^2}},$$ which splits into $${x+4y+8z+50 \over \sqrt{1^2+4^2+8^2}} = \pm{3x+4y+12z+82 \over \sqrt{3^2+4^2+12^2}}.$$ You can cross-multiply to clear the denominators just like we did when computing the angle bisectors of the normals.