\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Vector Lines-Planes Application - Intersections Involving Points, Lines and Planes

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On this page we cover applications involving intersections of points, lines and planes. Here is a list of topics that we cover.

Parallel Planes Through Points

Intersection of a Line and Plane

Line of Intersection of Two Planes

Parallel Planes Through Point

The simplest problems involve finding parallel lines through points. Here is a quick practice problem to get you started.

Practice

Find a vector equation for the line that goes through the point \( (1,0,-3) \) and is parallel to \( \vec{v} = 2\hat{i} - 4\hat{j} + 5\hat{k} \).

Problem Statement

Find a vector equation for the line that goes through the point \( (1,0,-3) \) and is parallel to \( \vec{v} = 2\hat{i} - 4\hat{j} + 5\hat{k} \).

Solution

1250 video

video by Dr Chris Tisdell

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Intersection of a Line and Plane

Here is a good video clip explaining how to find the intersection of a line and a plane.

PatrickJMT - intersection of line and plane [1min-1sec]

video by PatrickJMT

Practice

Find the point at which the line given by the points \( (1,0,1) \) and \( (4,-2,2) \) intersects the plane \( x + y + z = 6 \).

Problem Statement

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

1253 video

video by PatrickJMT

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Find the intersection of the plane \( 2x -8y + z = -77 \) and the line \( \vec{r}(t) = \langle -1,2,-6 \rangle t + \langle -1,6,-3\rangle \).

Problem Statement

Find the intersection of the plane \( 2x -8y + z = -77 \) and the line \( \vec{r}(t) = \langle -1,2,-6 \rangle t + \langle -1,6,-3\rangle \).

Solution

2297 video

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Line of Intersection of Two Planes

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

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

Find the line of intersection of the planes \( 3x + 2y - z = 7 \) and \( x - 3y + 2z = 0 \) using the column elimination technique.

Problem Statement

Find the line of intersection of the planes \( 3x + 2y - z = 7 \) and \( x - 3y + 2z = 0 \) using the column elimination technique.

Final Answer

Parametric Equations: \( x = t; ~~~ y = 14 - 7t; ~~~ z = 21 - 11t \)

Problem Statement

Find the line of intersection of the planes \( 3x + 2y - z = 7 \) and \( x - 3y + 2z = 0 \) using the column elimination technique.

Solution

Multiply the first equation by \(2\) and add it to the second equation. This will eliminate the z component.
\(\begin{array}{rcrcrcl} 3x & + & 2y & - & z & = & 7 \\ x & - & 3y & + & 2z & = & 0 \end{array} \) \( ~~~ \to ~~~ \) \(\begin{array}{rcrcrcl} 6x & + & 4y & - & 2z & = & 14 \\ x & - & 3y & + & 2z & = & 0 \end{array} \)
Adding the 2 equations in the column on the right above yields \( ~~ 7x +y = 14 ~~ \to ~~ y = 14 - 7x \)
We can now substitute \( y = 14 - 7x \) into one of the original equations. It doesn't matter which one we choose. For this example, we will use the second one, \( x - 3y + 2z = 0 \).
\(\begin{array}{rcl} x - 3(14-7x) + 2z & = & 0 \\ x - 42 + 21x + 2z & = & 0 \\ 22x + 2z & = & 42 \\ 11x + z & = & 21 \\ z & = & 21 - 11x \end{array}\)
So now we have the equations for \(y\) and \(z\) in terms of \(x\). There are many (in fact, an infinite number) choices for the parameter \(t\). For this example, we will choose the most obvious, \( x = t \). This gives us the parametric equations
\(\begin{array}{rcl} x & = & t \\ y & = & 14 - 7t \\ z & = & 21 - 11t \end{array}\)

Final Answer

Parametric Equations: \( x = t; ~~~ y = 14 - 7t; ~~~ z = 21 - 11t \)

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Technique 2: Cross Product of Normal Vectors

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

Find the line of intersection of the planes \(3x+2y-z=7\) and \(x-3y+2z=0\) using the cross product technique.

Problem Statement

Find the line of intersection of the planes \(3x+2y-z=7\) and \(x-3y+2z=0\) using the cross product technique.

Final Answer

Point: \( (2, 0, -1)\)
Direction Vector: \( \hat{i} - 7 \hat{j} - 11 \hat{k} \)

Problem Statement

Find the line of intersection of the planes \(3x+2y-z=7\) and \(x-3y+2z=0\) using the cross product technique.

Solution

Get the normal vectors from the equations of the lines.
\( 3x + 2y - z = 7 ~~ \to ~~ \vec{n}_1 = 3\hat{i} + 2\hat{j} - \hat{k} \)
\( x - 3y + 2z = 0 ~~ \to ~~ \vec{n}_2 = \hat{i} -3\hat{j} + 2\hat{k} \)
Calculate the cross product.
\( \displaystyle{\vec{n}_1 \times \vec{n}_2 = }\) \(\displaystyle{\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 1 & -3 & 2 \end{vmatrix} = }\) \(\displaystyle{ \hat{i} - 7 \hat{j} - 11 \hat{k}}\)
Notice that the vector components are the same as the direction numbers of the parametric equations that we found in the technique 1 solution, i.e. \( a=1, b=-7, c=-11 \)
So, now we have a vector in the direction of the line, we need one point to determine the line. Since the line is the intersection of the two planes, we just need to find one point that is on both planes. We have two equations and three variables. So if we choose one the components, we can solve the remaining equations to get the other two components.
The easiest is to choose zero for the most complicated component. So let's let \(y=0\) and solve for x and z.
\( 3x +2(0) - z = 7 ~~ \to ~~ 3x - z = 7 \)
\( x - 3(0) + 2z = 0 ~~ \to ~~ x + 2z = 0 \)
Multiply the first equation by \(2\) and add to the second to get
\( 6x - 2z = 14 \)
\( 7x = 14 ~~ \to ~~ x = 2 \)
Substituting \(x=2\) into the second equation yields \( 2 + 2z = 0 ~~ \to ~~ z = -1 \)

Final Answer

Point: \( (2, 0, -1)\)
Direction Vector: \( \hat{i} - 7 \hat{j} - 11 \hat{k} \)

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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 \( 2x + 3y + z = 16 \).

Problem Statement

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

1766 video

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Really UNDERSTAND Calculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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