Specifying one point \((x_0,y_0,z_0)\) on a plane and a vector \(\textbf
\[\begin \textbf\cdot\left \langle x-x_0,y-y_0,z-z_0 \right \rangle =0 \end\] Writing out in components \[\begin n_x(x-x_0)+n_y(y-y_0)+n_z(z-z_0)=0 \quad\text\quad n_xx+n_yy+n_zz= d \end\] where \(d=n_xx_0+n_yy_0+n_zz_0\text<.>\)
Again, the coefficients \(n_x,n_y,n_z\) of \(x,\ y\) and \(z\) in the equation of the plane are the components of a vector \(\left \langle n_x,n_y,n_z \right \rangle \) perpendicular to the plane. The vector \(\textbf
We have just seen that if we write the equation of a plane in the standard form \[ ax+by+cz=d \nonumber \] then it is easy to read off a normal vector for the plane. It is just \(\left \langle a,b,c \right \rangle\text<.>\) So for example the planes \[ P:\ x+2y+3z=4 \qquad P':\ 3x+6y+9z=7 \nonumber \] have normal vectors \(\textbf=\left \langle 1,2,3 \right \rangle\) and \(\textbf'=\left \langle 3,6,9 \right \rangle\text\) respectively. Since \(\textbf'=3\textbf\text\) the two normal vectors \(\textbf\) and \(\textbf'\) are parallel to each other. This tells us that the planes \(P\) and \(P'\) are parallel to each other. When the normal vectors of two planes are perpendicular to each other, we say that the planes are perpendicular to each other. For example the planes \[ P:\ x+2y+3z=4 \qquad P'':\ 2x-y=7 \nonumber \] have normal vectors \(\textbf=\left \langle 1,2,3 \right \rangle\) and \(\textbf''=\left \langle 2,-1,0 \right \rangle\text\) respectively. Since \[ \textbf\cdot\textbf'' = 1\times 2+2\times(-1)+3\times 0 = 0 \nonumber \] the normal vectors \(\textbf\) and \(\textbf''\) are mutually perpendicular, so the corresponding planes \(P\) and \(P''\) are perpendicular to each other.
Here is an example that illustrates how one can sketch a plane, given the equation of the plane.That's what we'll do. The intersection of the plane \(P\) with the \(xy\)-plane is the straight line through the two points \((3,0,0)\) and \((0,4,0)\text<.>\) So the part of that intersection in the first octant is the line segment from \((3,0,0)\) to \((0,4,0)\text<.>\) Similarly the part of the intersection of \(P\) with the \(xz\)-plane that is in the first octant is the line segment from \((3,0,0)\) to \((0,0,6)\) and the part of the intersection of \(P\) with the \(yz\)-plane that is in the first octant is the line segment from \((0,4,0)\) to \((0,0,6)\text<.>\) So we just have to sketch the three line segments joining the three axis intercepts \((3,0,0)\text\) \((0,4,0)\) and \((0,0,6)\text<.>\) That's it.
Here are two examples that illustrate how one can find the distance between a point and a plane.
In this example, we'll compute the distance between the point
\[ \textbf = (1,-1,-3) \qquad\text\qquad P:\ x+2y+3z=18 \nonumber \]
By the “distance between \(\textbf\) and the plane \(P\)” we mean the shortest distance between \(\textbf\) and any point \(\textbf\) on \(P\text<.>\) In fact, we'll evaluate the distance in two different ways. In the next Example 1.4.5, we'll use projection. In this example, our strategy for finding the distance will be to
So imagine that we start walking, and that we start at time \(t=0\) at \(\textbf\) and walk in the direction \(\textbf\text<.>\) Then at time \(t\) we might be at
\[ \textbf+t\textbf = (1,-1,-3) +t\,\left \langle 1,2,3 \right \rangle = (1+t, -1+2t, -3+3t) \nonumber \]
We hit the plane \(P\) at exactly the time \(t\) for which \((1+t, -1+2t, -3+3t)\) satisfies the equation for \(P\text\) which is \(x+2y+3z=18\text<.>\) So we are on \(P\) at the unique time \(t\) obeying
\[\begin (1+t)+2(-1+2t)+3(-3+3t)=18 &\iff 14t = 28 \iff t=2 \end\]
So the point on \(P\) which is closest to \(\textbf\) is
and the distance from \(\textbf\) to \(P\) is the distance from \(\textbf\) to \(\textbf\text\) which is
We are again going to find the distance from the point
\[ \textbf = (1,-1,-3) \qquad\text\qquad P:\ x+2y+3z=18 \nonumber \]
But this time we will use the following strategy.
Now let's find a point on \(P\text<.>\) The plane \(P\) is given by a single equation, namely
in the three unknowns, \(x\text\) \(y\text\) \(z\text<.>\) The easiest way to find one solution to this equation is to assign two of the unknowns the value zero and then solve for the third unknown. For example, if we set \(x=y=0\text\) then the equation reduces to \(3z=18\text<.>\) So we may take \(\textbf=(0,0,6)\text<.>\)
Then \(\textbf\text\) the vector from \(\textbf=(1,-1,-3)\) to \(\textbf=(0,0,6)\) is \(\left \langle 0-1\,,\,0-(-1)\,,\,6-(-3) \right \rangle=\left \langle -1,1,9 \right \rangle\) so that, by Equation 1.2.14,
and the distance from \(\textbf\) to \(P\) is
\[\begin \left|_>\,\textbf\right| = \big|2 \left \langle 1,2,3 \right \rangle\big| =2\sqrt \end\]
just as we found in Example 1.4.4.
In the next example, we find the distance between two planes.
Now we'll increase the degree of difficulty a tiny bit, and compute the distance between the planes
\[ P:\ x+2y+2z=1 \qquad\text\qquad P':\ 2x+4y+4z=11 \nonumber \]
By the “distance between the planes \(P\) and \(P'\)” we mean the shortest distance between any pair of points \(\textbf\) and \(\textbf'\) with \(\textbf\) in \(P\) and \(\textbf'\) in \(P'\text<.>\) First observe that the normal vectors
\[ \textbf=\left \langle 1,2,2 \right \rangle \qquad\text\qquad \textbf'=\left \langle 2,4,4 \right \rangle=2\textbf \nonumber \]
to \(P\) and \(P'\) are parallel to each other. So the planes \(P\) and \(P'\) are parallel to each other. If they had not been parallel, they would have crossed and the distance between them would have been zero.
Our strategy for finding the distance will be to
We can find a point on \(P\) just as we did on Example 1.4.5. The plane \(P\) is given by the single equation
in the three unknowns, \(x\text\) \(y\text\) \(z\text<.>\) We can find one solution to this equation by assigning two of the unknowns the value zero and then solving for the third unknown. For example, if we set \(y=z=0\text\) then the equation reduces to \(x=1\text<.>\) So we may take \(\textbf=(1,0,0)\text<.>\)
Now imagine that we start walking, and that we start at time \(t=0\) at \(\textbf\) and walk in the direction \(\textbf\text<.>\) Then at time \(t\) we might be at
\[ \textbf+t\textbf = (1,0,0) +t\left \langle 1,2,2 \right \rangle = (1+t, 2t, 2t) \nonumber \]
We hit the second plane \(P'\) at exactly the time \(t\) for which \((1+t, 2t, 2t)\) satisfies the equation for \(P'\text\) which is \(2x+4y+4z=11\text<.>\) So we are on \(P'\) at the unique time \(t\) obeying
So the point on \(P'\) which is closest to \(\textbf\) is
and the distance from \(P\) to \(P'\) is the distance from \(\textbf\) to \(\textbf'\) which is
Now we'll find the angle between two intersecting planes.
The orientation (i.e. direction) of a plane is determined by its normal vector. So, by definition, the angle between two planes is the angle between their normal vectors. For example, the normal vectors of the two planes
\[\begin \textbf_1&=\left \langle 2,1,-1 \right \rangle\\ \textbf_2&=\left \langle 1,1,1 \right \rangle \end\]
If we use \(\theta\) to denote the angle between \(\textbf_1\) and \(\textbf_2\text\) then
to four decimal places. That's in radians. In degrees, it is \(1.0799\frac<\pi>=61.87^\circ\) to two decimal places.
