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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.

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.

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 - Intersections Involving Points, Lines and Planes

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

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 solution video

video by PatrickJMT

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Finding The 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

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 solution 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 solution video

video by MIP4U

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Finding The 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.

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.

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: \( \hat{i} - 7 \hat{j} - 11 \hat{k} \)

close solution

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 solution video

video by Dr Chris Tisdell

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Section 2 - Calculating Distances Between Points, Lines and Planes

Distance Between A Point and A Line In Space

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)

video by Dr Chris Tisdell

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) [1min-17secs]

video by MIP4U

Okay, now let's work some practice problems calculating distances between points and lines.

Determine the distance between the line \( x=2+3t, y=-4+t, z=-2t \) and the point \( (1,-6,2) \).

Problem Statement

Determine the distance between the line \( x=2+3t, y=-4+t, z=-2t \) and the point \( (1,-6,2) \).

Solution

1848 solution video

video by MIP4U

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Calculate the distance between the point \( B(1,2,3) \) and the line \(x = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix}\)

Problem Statement

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

1856 solution video

video by Dr Chris Tisdell

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Find the distance between the point \( (4,1,-2) \) and the line \( x=1+t, y=3-2t, z=4-3t \).

Problem Statement

Find the distance between the point \( (4,1,-2) \) and the line \( x=1+t, y=3-2t, z=4-3t \).

Solution

1857 solution video

video by Krista King Math

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Find the distance between the point \( Q(4,4,1) \) and the line \( x=2t+1, y=-t+7, z=8t+5 \).

Problem Statement

Find the distance between the point \( Q(4,4,1) \) and the line \( x=2t+1, y=-t+7, z=8t+5 \).

Solution

1862 solution video

video by Firefly

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Distance Between Two Parallel Lines

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

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.

Compute the distance between the point \( B(1,2,3) \) and the plane \( x = \begin{bmatrix} 5 \\ 0 \\ 1 \end{bmatrix} + \lambda_1 \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix} + \lambda_2 \begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix} \)

Problem Statement

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

1854 solution video

video by Dr Chris Tisdell

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Find the distance between the point \( B(1,2,3) \) and the plane \( 2x - y + z = 5 \)

Problem Statement

Find the distance between the point \( B(1,2,3) \) and the plane \( 2x - y + z = 5 \)

Solution

1855 solution video

video by Dr Chris Tisdell

close solution

Find the distance between the point \( (1,-2,4) \) and the plane \( 3x + 2y + 6z = 5 \).

Problem Statement

Find the distance between the point \( (1,-2,4) \) and the plane \( 3x + 2y + 6z = 5 \).

Solution

1859 solution video

video by Krista King Math

close solution

Find the distance between the point \( (7,-3,9) \) and the plane \( x - 2y + 3z = 6 \).

Problem Statement

Find the distance between the point \( (7,-3,9) \) and the plane \( x - 2y + 3z = 6 \).

Solution

1860 solution video

video by MIP4U

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Find the distance between the point \( Q(3,8,-1) \) and the plane \( 4(x+2) - (y+5) + 2z = 0 \).

Problem Statement

Find the distance between the point \( Q(3,8,-1) \) and the plane \( 4(x+2) - (y+5) + 2z = 0 \).

Solution

1863 solution video

video by Firefly

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Find the distance between the point \( (0,0,0) \) and the plane \( 2x + y - 2z = 4 \).

Problem Statement

Find the distance between the point \( (0,0,0) \) and the plane \( 2x + y - 2z = 4 \).

Solution

1864 solution video

video by MIT OCW

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Distance Between Two Parallel Planes

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.

Find the distance between the parallel planes \( 2x - 3y + z = 4 \) and \( 4x - 6y + 2z = 3 \).

Problem Statement

Find the distance between the parallel planes \( 2x - 3y + z = 4 \) and \( 4x - 6y + 2z = 3 \).

Solution

1858 solution video

video by Krista King Math

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Find the distance between the parallel planes \( -5x + 3y - 2z = -14 \) and \( -5x + 3y - 2z = -25 \).

Problem Statement

Find the distance between the parallel planes \( -5x + 3y - 2z = -14 \) and \( -5x + 3y - 2z = -25 \).

Solution

1861 solution video

video by MIP4U

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Section 3 - Tangents and Normals to Planes

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 \( (3,3,5) \).

Problem Statement

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

1482 solution video

video by Krista King Math

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Determine the equation of the line that passes through the point \( (1,-1,1) \) and is normal to the plane \( 2x + 3y - z = 4 \).

Problem Statement

Determine the equation of the line that passes through the point \( (1,-1,1) \) and is normal to the plane \( 2x + 3y - z = 4 \).

Solution

1767 solution video

video by Dr Chris Tisdell

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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 \( A(2,-1,6) \).

Problem Statement

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

1828 solution video

video by Dr Chris Tisdell

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