This kind:
In order to do anything with planes we need a standard equation for them. For example...
This:
Ax+By+Cz+D=0
But let's face it, although we can use that equation to figure out the x, y and z-intercepts, and use it to find intercepts with other planes etc. we can't really use that form to find an equation for a specific plane.
To find the equation of a plane we need two things, a point on said plane, and a vector that is perpendicular to everything on the plane or, more simply that is normal to it.
So we have point (xo, yo, zo) in the plane and vector <A, B, C> which is normal to the plane. How do we take these and make them into an equation for a plane? Simple! We know that a ∙ b=0 if a and b are perpendicular so if we say that point (x, y, z) is also on the plane we can say that <x-xo, y-yo, z-zo> is also on the plane. Meaning:
<A, B, C> ∙ <x-xo, y-yo, z-zo>=0
We can solve this dot product to get:
A(x-xo) + B(y-yo) + C(z-zo)=0
Which is the point normal form of a plane.
To break it down, A, B and C are the x, y, and z coordinates of a vector normal to a plane. xo, yo, and zo are the x, y and z coordinates of a point on the plane and, x,y, and z are the variables in the equation.
IMPORTANT: In standard form the plane equation ALWAYS equals zero, if it doesn't it's not in standard form. Always keep track of the constant D value. Ye have been warned.
Armed with the knowledge of plane equations we can do all sorts of stuff with planes like find out if two planes are parallel, perpendicular, their line of intersection, their angle of intersection and even the distance between two planes!
Lets start out simple, determining if two planes are parallel or perpendicular.
If two planes are parallel, their normals are scalar multiples. For example planes with normals <a,b,c> and <ak,bk,ck> are parallel planes because one of the normals is simply the other multiplied by scalar k.
Can't tell at a glance if two vectors are scalar multiples? If a/a1= b/b1 = c/c1 then they're parallel.
Two planes are perpendicular if the dot product of their normals is zero.
Simple enough right?
Now adding lines into the mix doesn't make anything harder but it is very important to remember that the vector given in the line equation is parallel to the line not perpendicular so to find if a line is parallel to a plane the dot product of the vector slope of the line and the normal vector must be zero and if they are perpendicular the vectors will be scalar multiples.
One other little tidbit before moving on, if a point is plugged into a plane equation in standard or point-normal form and it is equal to zero the point is on the plane, if it's positive it's above the plane and if it's negative it's below the plane.
Now we know how to tell if two planes are parallel and where a point is in relation to a plane, what if we want to solve for the line where two planes intersect?
This is actually quite simple to do but the logic and reasoning behind it can be a little hard to grasp at first. Say we have two planes x+2y+3z=0 and 3y+4z=0 and want to find their line of intersection. The first thing we need to do is recall the standard equation for a line in 3-space <x, y, z>=<xo, yo, zo>+t<a,b,c> and that it requires a point (xo, yo, zo) for and intercept and a vector <a, b, c> for a slope. The point (xo, yo, zo) we can find by finding a point that is in both our planes like (0, 0, 0). Now that we have a point we need to find the slope, this can seem difficult at first but the slope of this line is actually the cross product of the normals of the intersecting planes. Each normal is perpendicular to its respective plane so the vector perpendicular to those must be parallel to both the planes, in other words have the same slope of the line of intersection.
Find the cross product, which in this case is <-1, -4, 3> and plug the point and vector into the line equation to get <x, y, z>=<0, 0, 0> +t<-1, -4, 3>. That;s our line on the intersection of the planes x+2y+3z=0 and 3y+4z=0!
The key step when it comes to finding the angle between planes is recognizing that the angle we want is also the angle between the normals of the two planes, which are simply the coefficients of the equations for the planes.
This means that we can the find the angle between the two normals using the formula from section 13-4, cos(θ)=aŸb/||a||||b||.
Just don’t forget that the angle we want to find is always the acute angle between the planes, so if the angle found from the above equation is obtuse, more than 90˚, then we just need to subtract that value from 180˚ to find it’s supplementary angle.
Distance formula!!!
The distance D between the point P(x0,y0,z0) and the plane ax+by+cz+d=0 is D=|ax0+by0+cz0+d|/√(a2+b2+c2)
Now for the tricky part: If the question asks for the distance between two lines, L1 and L2 for instance, it gets a little more complicated.
First, find a point on L1 by setting t equal to 0. We’ll use this point as P in the distance formula.
Next, find the point normal form of the plane containing L2 by finding the cross product of L1 and L2, which is the normal to the lines, and a point on L2, we’ll call it <x1,y1,z1>, by setting t equal to 0. The point normal form is equal to the normal dotted with the vector from the point we found to the origin, or n1(x-x1)+n2(y-y1)+ n3(z-z1).
Once we have a point on L1 and a plane containing L2, we can solve for the distance between them using the distance formula.
This strategy works for any combination of points, lines, and planes. If we’re given a line and a plane for instance, we just need to find a point on the line by setting t equal to 0, and then we can use the distance formula. As long as we end up with a point and a plane, we’re golden.
By Katelyn and Alice
We can solve this dot product to get:
A(x-xo) + B(y-yo) + C(z-zo)=0
Which is the point normal form of a plane.
To break it down, A, B and C are the x, y, and z coordinates of a vector normal to a plane. xo, yo, and zo are the x, y and z coordinates of a point on the plane and, x,y, and z are the variables in the equation.
IMPORTANT: In standard form the plane equation ALWAYS equals zero, if it doesn't it's not in standard form. Always keep track of the constant D value. Ye have been warned.
Armed with the knowledge of plane equations we can do all sorts of stuff with planes like find out if two planes are parallel, perpendicular, their line of intersection, their angle of intersection and even the distance between two planes!
Lets start out simple, determining if two planes are parallel or perpendicular.
If two planes are parallel, their normals are scalar multiples. For example planes with normals <a,b,c> and <ak,bk,ck> are parallel planes because one of the normals is simply the other multiplied by scalar k.
Can't tell at a glance if two vectors are scalar multiples? If a/a1= b/b1 = c/c1 then they're parallel.
Two planes are perpendicular if the dot product of their normals is zero.
Simple enough right?
Now adding lines into the mix doesn't make anything harder but it is very important to remember that the vector given in the line equation is parallel to the line not perpendicular so to find if a line is parallel to a plane the dot product of the vector slope of the line and the normal vector must be zero and if they are perpendicular the vectors will be scalar multiples.
One other little tidbit before moving on, if a point is plugged into a plane equation in standard or point-normal form and it is equal to zero the point is on the plane, if it's positive it's above the plane and if it's negative it's below the plane.
Now we know how to tell if two planes are parallel and where a point is in relation to a plane, what if we want to solve for the line where two planes intersect?
This is actually quite simple to do but the logic and reasoning behind it can be a little hard to grasp at first. Say we have two planes x+2y+3z=0 and 3y+4z=0 and want to find their line of intersection. The first thing we need to do is recall the standard equation for a line in 3-space <x, y, z>=<xo, yo, zo>+t<a,b,c> and that it requires a point (xo, yo, zo) for and intercept and a vector <a, b, c> for a slope. The point (xo, yo, zo) we can find by finding a point that is in both our planes like (0, 0, 0). Now that we have a point we need to find the slope, this can seem difficult at first but the slope of this line is actually the cross product of the normals of the intersecting planes. Each normal is perpendicular to its respective plane so the vector perpendicular to those must be parallel to both the planes, in other words have the same slope of the line of intersection.
Find the cross product, which in this case is <-1, -4, 3> and plug the point and vector into the line equation to get <x, y, z>=<0, 0, 0> +t<-1, -4, 3>. That;s our line on the intersection of the planes x+2y+3z=0 and 3y+4z=0!
This means that we can the find the angle between the two normals using the formula from section 13-4, cos(θ)=aŸb/||a||||b||.
Just don’t forget that the angle we want to find is always the acute angle between the planes, so if the angle found from the above equation is obtuse, more than 90˚, then we just need to subtract that value from 180˚ to find it’s supplementary angle.
Distance formula!!!
The distance D between the point P(x0,y0,z0) and the plane ax+by+cz+d=0 is D=|ax0+by0+cz0+d|/√(a2+b2+c2)
Now for the tricky part: If the question asks for the distance between two lines, L1 and L2 for instance, it gets a little more complicated.
First, find a point on L1 by setting t equal to 0. We’ll use this point as P in the distance formula.
Next, find the point normal form of the plane containing L2 by finding the cross product of L1 and L2, which is the normal to the lines, and a point on L2, we’ll call it <x1,y1,z1>, by setting t equal to 0. The point normal form is equal to the normal dotted with the vector from the point we found to the origin, or n1(x-x1)+n2(y-y1)+ n3(z-z1).
Once we have a point on L1 and a plane containing L2, we can solve for the distance between them using the distance formula.
This strategy works for any combination of points, lines, and planes. If we’re given a line and a plane for instance, we just need to find a point on the line by setting t equal to 0, and then we can use the distance formula. As long as we end up with a point and a plane, we’re golden.
By Katelyn and Alice
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