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Applications of Lines and Planes in 3-Space Using Vectors |
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Before reading this page, you need to understand the basics of lines in 3-space and planes in 3-space. This page covers topics combining lines and planes and is divided into three main sections. |
Section 1 - Finding Intersections Involving Lines and Planes |
Section 2 - Calculating Distances Between Points, Lines and Planes |
Some Concepts You Need To Know To Work Application Problems - - There are just too many applications of lines and planes in space to cover all of them here. So this section lists some of the concepts you need to know to work application problems and then shows some examples of how to apply them. The idea of application problems is to put these concepts together to solve the problem.
1. The dot product of orthogonal vectors is a scalar zero. An example where we use this concept is in the proof of the standard equation of a plane near the bottom of this page. |
2. The cross product of parallel vectors is a zero vector. |
3. The result of the cross product is a vector which is orthogonal to the original two vectors. |
4. Make sure you know how to convert from one form of a line or plane representation to any other. For example, if you are given a plane in parametric form, know how to get the general form. Working practice problems helps you think through how to do each case. |
Section 1 - Finding Intersections Involving Lines and Planes
Finding The Intersection of a Line and Plane |
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Here is a good video clip explaining how to find the intersection of a line and a plane.
PatrickJMT - intersection of line and plane | |
Practice 1 |
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Find the point at which the line given by the points \( (1,0,1) \) and \( (4,-2,2) \) intersects the plane \( x+y+z=6 \). |
solution |
Finding The Line of Intersection of Two Planes |
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We present two possible ways to determine the line of intersection of two planes. In both cases, we will be given the equations of the planes in general form. The technique to use depends on the problem statement and in what form your answer needs to be. In both cases there are an infinite number of solutions. We will discuss each technique and then explain how you can check to see if your answer is correct when it differs from a given answer.
Technique 1: Column Elimination to Get Parametric Equations |
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First, we can use column elimination and get the parametric equations of the line. The idea is the same as when you have two equations and two unknowns and you need solve a system of equations. The difference in this case is that we have two equations (the two planes) and three unknowns ( \( x,y,z \) ). So what we do is manipulate the equations to get two equations in terms of one of the variables. Then we assign the parameter to the remaining variable and we end up with three parametric equations.
As mentioned above, there are many (in fact, infinite) correct answers. So what do you do if your answer doesn't match the answer given? Does this mean your answer is incorrect? Not necessarily. Remember a line is uniquely defined by two points, any two points on the line. For this discussion, we will assume the answer is \( x = f(t); y = g(t); z = h(t) \). Choose any value for t. Let's call it \(t_1\). Get a point by setting \( t = t_1 \) in the parametric equations.
\( ( x_1, y_1, z_1 ) = ( f(t_1), g(t_1), h(t_1) ) \)
Now take each component and plug them into your parametric equations. Solve for t in all three cases. If the value of t is the same for all three, then that point is also on your line.
Now you need to repeat this procedure for a second point. Choose a different value for t, call it \(t_2\). If you solve for a different t in your parametric equations than you got for the first point, then you can guarantee that your answer is correct.
Question: Why do this twice? Wouldn't doing it once be enough to prove that the answers define the same line?
No. The first point just guarantees that the lines intersect and that may be the only point where the intersection occurs. Checking two values guarantees that the two sets of parametric equations define the same line.
Practice 2 | |
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Find the line of intersection of the planes \( 3x + 2y - z = 7 \) and \( x - 3y + 2z = 0 \) using the column elimination technique. | |
answer |
solution |
Technique 2: Cross Product of Normal Vectors |
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Second, we can use the normal vectors and take the cross product to get the normal vector. Then we find one point that is on both planes giving a point and a direction vector. These will uniquely define a line. But why does this work? Let's think about it.
If we have a normal vector to a plane, all lines in the plane are orthogonal to the normal vector. Also, when we take the cross product of two vectors, the result is a vector that is orthonal to both of the original vectors. So the result of the cross product is a vector that lies in both planes. Cool, eh?
So how do we find a point? We need to solve both of the original plane equations. The idea is that we need a point that is in both planes. The only place this will occur is on the line of intersection of the planes. We are given two equations (planes) and three unknowns ( \( x,y,z\) ). What we do is choose a convenient value for one of the unknowns (any value will do) and solve the remaining system of equations.
Similar to the idea in technique 1 above, there are infinitely many answers. So how do you know if your answer is correct? First, make sure your direction vector is parallel to the one given in the answer. If it is, check that your point satisfies both of the plane equations (plug the point into both equations). If both of these tests hold, then you are assured that your answer is correct.
Practice 3 | |
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Find the line of intersection of the planes \(3x+2y-z=7\) and \(x-3y+2z=0\) using the cross product technique. | |
answer |
solution |
Practice 4 |
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Determine the point of intersection of the line \(\langle 2,3,0 \rangle + \mu\langle 1,-1,4 \rangle\) with the plane \(2x+3y+z=16\). |
solution |
Okay, now let's look at finding distances between points, lines and planes.
Section 2 - Calculating Distances Between Points, Lines and Planes
Distance Between A Point and A Line In Space |
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Using the figure on the right, we will show how to find the distance d between the line L and the point Q. There are several ways to do this. We will look at two in particular, one using the dot product, the other using the cross product.
Using the dot product:
1. Find a vector along the line L. This vector does not necessarily have to be a unit vector.
2. Find a vector from any point P on the line to the point Q.
3. Calculate the projection of the vector \( \overrightarrow{PQ}\) onto \(\vec{u}\), \(proj_{\vec{u}}\overrightarrow{PQ}\).
4. Find \( \overrightarrow{PQ} - proj_{\vec{u}}\overrightarrow{PQ} \)
5. Calculate \( d = \| \overrightarrow{PQ} - proj_{\vec{u}}\overrightarrow{PQ} \| \)
Here is a great video explaining this in more detail and showing an example.
Dr Chris Tisdell - Distance from point to line (using the dot product) | |
Using the cross product:
The idea is to determine a point on the line and call it P (Any point will do) which we use to find the vector \(\overrightarrow{PQ}\). Then we find a unit vector \(\vec{u}\) along the line.
Now we can use \(\vec{u}\) and \( \overrightarrow{PQ}\) to determine the sine of the angle \(\theta\) as \(\displaystyle{
\sin(\theta) = \frac{d}{\| \overrightarrow{PQ} \|} ~~~ \to ~~~ d = \| \overrightarrow{PQ} \| \sin(\theta) }\)
From a geometric property of the cross product we know that
\(\displaystyle{ \| \overrightarrow{PQ} \times \vec{u} \| = \| \overrightarrow{PQ} \| ~ \| \vec{u} \| sin(\theta) }\) where \( \theta \) is the angle between the vectors. So \(\displaystyle{ sin(\theta) = \frac{\| \overrightarrow{PQ} \times \vec{u} \|}{\| \overrightarrow{PQ} \| ~ \| \vec{u} \|} }\). Substituting \(sin(\theta) \) from this equation into the equation for d we get
\(\displaystyle{ d = \| \overrightarrow{PQ} \| \frac{\| \overrightarrow{PQ} \times \vec{u} \|}{\| \overrightarrow{PQ} \| ~ \| \vec{u} \|} = \frac{\| \overrightarrow{PQ} \times \vec{u} \|}{ \| \vec{u} \|}. }\)
And since \(\vec{u}\) is a unit vector, it's length is one, so we can write the equation for \(d \) as \(\displaystyle{ d = \| \overrightarrow{PQ} \times \vec{u} \|. }\)
Here is a video clip explaining this derivation.
MIP4U - Determining the Distance Between a Line and a Point (using the cross product) | |
Okay, now let's work some practice problems calculating distances between points and lines.
Practice 5 |
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Determine the distance between the line \(x=2+3t, y=-4+t, z=-2t\) and the point \((1,-6,2)\). |
solution |
Practice 6 |
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Calculate the distance between the point \(B(1,2,3)\) and the line \( x=\begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix}\) |
solution |
Practice 7 |
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Find the distance between the point \((4,1,-2)\) and the line \(x=1+t, \) \(y=3-2t, \) \(z=4-3t\). |
solution |
Practice 8 |
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Find the distance between the point \(Q(4,4,1)\) and the line \(x=2t+1, y=-t+7, z=8t+5\). |
solution |
Distance Between Two Parallel Lines |
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To find the distance between two parallel lines, we extend the logic of the previous section by first finding one point on one of the lines (any point will do) and then finding the distance between that point and the other line.
Distance Between a Point and a Plane |
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The goal is to find the distance between a point in space and a plane. In the figure on the right, we want to find the distance d, the perpendicular distance between Q and the plane.
As an overview, here is how we will do this. We will find a unit normal vector to the plane at some point P. We then project the vector \( \overrightarrow{PQ} \) onto the unit normal vector, labeled \(proj\) in the figure on the right. This distance is equal to the distance d. Spend a few minutes going over this discussion and looking at the figure to make sure you understand this.
We use the dot product to get the projection of one vector onto another. The equation is \(\displaystyle{ d = \left| \overrightarrow{PQ} \cdot \frac{\vec{n}}{\| \vec{n} \|} \right| = \frac{\left| \overrightarrow{PQ} \cdot \vec{n} \right|}{\| \vec{n} \|} }\)
Okay, time for some practice problems calculating the distance between a point and a plane.
Practice 9 |
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Compute the distance between the point \(B(1,2,3)\) and the plane \( x=\begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda_1 \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix} + \lambda_2 \begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix} \) |
solution |
Practice 10 |
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Find the distance between the point \(B(1,2,3)\) and the plane \(2x-y+z=5\) |
solution |
Practice 11 |
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Find the distance between the point \((1,-2,4)\) and the plane \(3x+2y+6z=5\). |
solution |
Practice 12 |
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Find the distance between the point \((7,-3,9)\) and the plane \(x-2y+3z=6\). |
solution |
Practice 13 |
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Find the distance between the point \(Q(3,8,-1)\) and the plane \(4(x+2) - (y+5)+2z=0\). |
solution |
Practice 14 |
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Find the distance between the point \((0,0,0)\) and the plane \(2x+y-2z=4\). |
solution |
Distance Between Two Parallel Planes |
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To find the distance between two parallel planes, we extend the logic of the previous section by first finding one point on one of the planes (any point will do) and then finding the distance between that point and the other plane.
Practice 15 |
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Find the distance between the parallel planes \(2x-3y+z=4\) and \(4x-6y+2z=3\). |
solution |
Practice 16 |
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Find the distance between the parallel planes \(-5x+3y-2z=-14\) and \(-5x+3y-2z=-25\). |
solution |
Section 3 - Other Applications
This section contains other types of applications involving points, lines and planes. These usually involve tangent planes and normal lines to a plane.
Practice 17 |
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Find the equation of the tangent plane and the symmetric equations of the normal line to the surface \(2(x-2)^2+(y-1)^2+(x-3)^2=10\) at the point \((3,3,5)\). |
solution |
Practice 18 |
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Determine the equation of the line that passes through \((1,-1,1)\) and normal to the plane \(2x+3y-z=4\). |
solution |
Practice 19 |
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Determine a normal vector and the equation of the tangent plane to the surface \(z=x^2+2y^2\) at the point \(A(2,-1,6)\). |
solution |