Three Dimensional Space Problems
Vector Problems

Let us continue with more examples.

Example 15.

Consider a line in 3-space

Choose a point on the line, call it \left (x_{0} ,y_{0} ,z_{0}\right ) . Our goal is to explain any other point on this line. Obviously, the yellow vector is \left \langle x_{0} ,y_{0} ,z_{0}\right \rangle . Now if \left (x ,y ,z\right ) ia any other point on the line, then the vector \left \langle x -x_{0} ,y -y_{0} ,z -z_{0}\right \rangle must be parallel to the line. We show this vector in red. Observe that for the vector \left \langle x ,y ,z\right \rangle which has its terminal point on the line at the point \left (x ,y ,z\right )

(1)   \begin{align*}\langle x ,y ,z\rangle =\langle x_{0} ,y_{0} ,z_{0}\rangle +\langle x -x_{0} ,y -y_{0} ,z -z_{0}\rangle \\ \,\end{align*}

This gives us a hint that if we have any vector \boldsymbol{v} =\left \langle a ,b ,c\right \rangle parallel to the line, with multiplying by the right parameter, we can get the vector \left \langle x ,y ,z\right \rangle . In fact

(2)   \begin{equation*}\left \langle x ,y ,z\right \rangle =\left \langle x_{0} ,y_{0} ,z_{0}\right \rangle +t\left \langle a ,b ,c\right \rangle \end{equation*}

This way, any vector with the initial point at the origin and the terminal point on the line can be explained. Observe that this information explains every point on the line. This form of equation of a line is called the vector form. One can obtain

(3)   \begin{align*}x =x_{0} +at \\ y =y_{0} +bt \\ z =z_{0} +ct\end{align*}

which is called the parametric form. Solving for t,

(4)   \begin{equation*}\frac{x -x_{0}}{a} =\frac{y -y_{0}}{b} =\frac{z -z_{0}}{c} \end{equation*}

which is called the symmetric form of equation of a line.

Example 16.

Consider a plane in 3-space.

Choose a point \left (x_{0} ,y_{0} ,z_{0}\right ) on this plane. Now we would like to explain any other point on this plane. Clearly, for any other point \left (x ,y ,z\right ) on the plane, the vector \left \langle x -x_{0} ,y -y_{0} ,z -z_{0}\right \rangle is on the plane. In the event that we have a vector \left \langle a ,b ,c\right \rangle perpendicular to the plane, we can write

(5)   \begin{equation*}\left \langle a ,b ,c\right \rangle \cdot \left \langle x -x_{0} ,y -y_{0} ,z -z_{0}\right \rangle =0 \end{equation*}

This choice of a perpendicular vector to the plane is called a normal to the plane and in the picture is shown in yellow. Obviously, a different choice of a perpendicular will not result on a different plane since the point is fixed on the plane to begin with. One can write the above plane equation in the form

(6)   \begin{equation*}a\left (x -x_{0}\right ) +b\left (y -y_{0}\right ) +c\left (z -z_{0}\right ) =0 \end{equation*}

or

(7)   \begin{equation*}ax +by +cz =ax_{0} +by_{0} +cz_{0} \end{equation*}

The right side of the above equation is a number.

Cross Product.

Suppose \boldsymbol{u} =\left \langle u_{1} ,u_{2} ,u_{3}\right \rangle and \boldsymbol{v} =\left \langle v_{1} ,v_{2} ,v_{3}\right \rangle are two vectors in 3-space. We define cross product of \boldsymbol{u} and \boldsymbol{v} by

(8)   \begin{align*}\boldsymbol{u} \times \boldsymbol{v} =\left \langle u_{1} ,u_{2} ,u_{3}\right \rangle \times \left \langle v_{1} ,v_{2} ,v_{3}\right \rangle \\ =\left \langle u_{2}v_{3} -u_{3}v_{2} ,u_{3}v_{1} -u_{1}v_{3} ,u_{1}v_{2} -u_{2}v_{1}\right \rangle \\ =\left (u_{3}v_{2} -v_{3}u_{2}\right )\boldsymbol{i} +\left (u_{1}v_{3} -v_{1}u_{3}\right )\boldsymbol{j} +\left (u_{2}v_{1} -v_{2}u_{1}\right )\boldsymbol{k}\end{align*}

Example 17.

We find cross product of a vector \langle a ,b ,c\rangle with itself

(9)   \begin{align*}\langle a ,b ,c\rangle \times \langle a ,b ,c\rangle =\langle bc -bc ,ac -ac ,ab -ab\rangle \\ =\langle 0 ,0 ,0\rangle \end{align*}

This is always true and note that the result of the cross product of two vectors is another vector. Many text books use the notion of determinant to compute the cross product of two vectors; here it is:

(10)   \begin{equation*}\left \langle u_{1} ,u_{2} ,u_{3}\right \rangle \times \left \langle v_{1} ,v_{2} ,v_{3}\right \rangle =\det \left (\begin{array}{ccc}\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3}\end{array}\right ) \end{equation*}

which essentially provides what we used as a definition. You can use this while computation but in this set of notes, we avoid it when possible.

A set of properties of cross product are as followed. Suppose k is a real number and \boldsymbol{u} ,\boldsymbol{v} ,\boldsymbol{w} are vectors in 3-space:

1. \boldsymbol{u} \times \boldsymbol{v} = -\boldsymbol{v} \times \boldsymbol{u}

2. k\left (\boldsymbol{u} \times \boldsymbol{v}\right ) =\left (k\boldsymbol{u}\right ) \times \boldsymbol{v} =\boldsymbol{u} \times \left (k\boldsymbol{v}\right )

3. \boldsymbol{u} \times \left (\boldsymbol{v} +\boldsymbol{w}\right ) =\left (\boldsymbol{u} \times \boldsymbol{v}\right ) +\left (\boldsymbol{u} \times \boldsymbol{w}\right )

Example 18.

We want to show

(11)   \begin{equation*}\left \Vert \boldsymbol{u} \times \boldsymbol{v}\right \Vert ^{2}=\Vert\boldsymbol{u}\Vert ^{2}\left \Vert \boldsymbol{v}\right \Vert ^{2} -\left (\boldsymbol{u} \cdot \boldsymbol{v}\right )^{2} \end{equation*}

Observe that on the left side

(12)   \begin{align*}\left \Vert \boldsymbol{u} \times \boldsymbol{v}\right \Vert ^{2} =(\boldsymbol{u} \times \boldsymbol{v}) \cdot (\boldsymbol{u} \times \boldsymbol{v}) \\ =\left (u_{3}v_{2} -v_{3}u_{2}\right )^{2} +\left (u_{2}v_{1} -v_{2}u_{1}\right )^{2} +\left (u_{1}v_{3} -v_{1}u_{3}\right )^{2} \\ \,\end{align*}

Now on the right side of the given expression (a seems to be an identity but we don’t know it yet!)

(13)   \begin{align*}\left \Vert \boldsymbol{u}\right \Vert ^{2}\left \Vert \boldsymbol{v}\right \Vert ^{2} -\left (\boldsymbol{u} \cdot \boldsymbol{v}\right )^{2} =(u_{1}^{2} +u_{2}^{2} +u_{3}^{2})(v_{1}^{2}+v_{2}^{2}+v_{3}^{2}) -\left (u_{1}v_{2} +u_{2}v_{2} +u_{3}v_{3}\right )^{2} \\ =u_{1}^{2}v_{1}^{2} +u_{1}^{2}v_{2}^{2} +u_{1}^{2}v_{3}^{2} +u_{2}^{2}v_{1}^{2} +u_{2}^{2}v_{2}^{2} +u_{2}^{2}v_{3}^{2} +u_{3}^{2}v_{1}^{2} +u_{3}^{2}v_{2}^{2} +u_{3}^{2}v_{3}^{2} \\ -\left (u_{1}^{2}v_{1}^{2} +u_{2}^{2}v_{2}^{2} +2u_{2}v_{2}u_{3}v_{3} +u_{3}^{2}v_{3}^{2} +2u_{1}v_{2}u_{2}v_{2} +2u_{1}v_{1}u_{3}v_{3}\right ) \\ =\left (u_{3}v_{2} -v_{3}u_{2}\right )^{2} +\left (u_{1}v_{3} -v_{1}u_{3}\right )^{2} +\left (u_{2}v_{1} -v_{2}u_{1}\right )^{2}\end{align*}

Therefore, the given statement is true. We are now done. Now consider 0 \leq \theta \leq \pi the angle between \boldsymbol{u} and \boldsymbol{v} . Another interesting relationship, is due to the fact that

(14)   \begin{equation*}\left (\boldsymbol{u} \cdot \boldsymbol{v}\right )^{2} =\left \Vert \boldsymbol{u}\right \Vert ^{2}\left \Vert \boldsymbol{v}\right \Vert ^{2}\cos ^{2}\theta =\left \Vert \boldsymbol{u}\right \Vert ^{2}\left \Vert \boldsymbol{v}\right \Vert ^{2}\left (1 -\sin ^{2}\theta \right ) \end{equation*}

and so

(15)   \begin{align*}\left \Vert \boldsymbol{u} \times \boldsymbol{v}\right \Vert ^{2} =\left \Vert \boldsymbol{u}\right \Vert ^{2}\left \Vert \boldsymbol{v}\right \Vert ^{2} -\left (\boldsymbol{u} \cdot \boldsymbol{v}\right )^{2} \\ =\left \Vert \boldsymbol{u}\right \Vert ^{2}\left \Vert \boldsymbol{v}\right \Vert ^{2} -\left \Vert \boldsymbol{u}\right \Vert ^{2}\left \Vert \boldsymbol{v}\right \Vert ^{2}\left (1 -\sin ^{2}\theta \right ) \\ =\left \Vert \boldsymbol{u}\right \Vert ^{2}\left \Vert \boldsymbol{v}\right \Vert ^{2}\sin ^{2}\theta \end{align*}

which provides

(16)   \begin{equation*}\left \Vert \boldsymbol{u} \times \boldsymbol{v}\right \Vert =\left \Vert \boldsymbol{u}\right \Vert \left \Vert \boldsymbol{v}\right \Vert \sin \theta \end{equation*}

Geometrically, we see that this is area of the shaded part that is a parallelogram with sides being vectors \boldsymbol{u} and \boldsymbol{v} . Recall that area of a parallelogram is obtained by multiplying base with hight.

Example 19.

Suppose \boldsymbol{u} and \boldsymbol{v} are vectors. We would like to find the angle between \boldsymbol{u} \times \boldsymbol{v} and \boldsymbol{u} . We know

(17)   \begin{equation*}\cos \theta =\frac{\left (\boldsymbol{u} \times \boldsymbol{v}\right ) \cdot \boldsymbol{u}}{\left \Vert \boldsymbol{u} \times \boldsymbol{v}\right \Vert \left \Vert \boldsymbol{u}\right \Vert }\end{equation*}

Computing the numerator we get

(18)   \begin{align*}\left (\boldsymbol{u} \times \boldsymbol{v}\right ) \cdot \boldsymbol{u} =\left \langle u_{2}v_{3} -u_{3}v_{2} ,u_{3}v_{1} -u_{1}v_{3} ,u_{1}v_{2} -u_{2}v_{1}\right \rangle \cdot \left \langle u_{1} ,u_{2} ,u_{3}\right \rangle \\ =u_{1}(u_{2}v_{3} -u_{3}v_{2}) +u_{2}\left (u_{3}v_{1} -u_{1}v_{3}\right ) +u_{3}\left (u_{1}v_{2} -u_{2}v_{1}\right ) =0\end{align*}

So the two vectors \boldsymbol{u} \times \boldsymbol{v} and \boldsymbol{u} are perpendicular to each other.

Example 20.

We want to write equation of a line that is parallel to the the vector \left \langle -1 ,1 ,2\right \rangle and passes the point \left (0 , -1 ,3\right ). We start by the vector form

(19)   \begin{equation*}\left \langle x ,y ,z\right \rangle =\left \langle 0 , -1 ,3\right \rangle +t\left \langle -1 ,1 ,2\right \rangle \end{equation*}

The parametric form of the equation is

(20)   \begin{equation*}x = -t ,y =t -1 ,z =2t +3 \end{equation*}

The symmetric form of the equation is

(21)   \begin{equation*}\frac{x -0}{ -1} =\frac{y -\left ( -1\right )}{1} =\frac{z -3}{2} \end{equation*}

Example 21.

Suppose we are given two lines

(22)   \begin{align*}l_{1} : -x =y +1 =\frac{z -3}{2} \\ l_{2} :\frac{x +2}{2} =\frac{y}{ -2} =\frac{z -1}{ -4}\end{align*}

We note that the line l_{1} is parallel to the vector \left \langle -1 ,1 ,2\right \rangle and l_{2} is parallel to the vector \left \langle 2 , -2 , -4\right \rangle. Since

(23)   \begin{equation*}\left \langle -1 ,1 ,2\right \rangle = -\frac{1}{2}\left \langle 2 , -2 , -4\right \rangle \end{equation*}

the two vectors are parallel to each other. So the two lines are parallel to one another.

Example 22.

Suppose we are given the two lines

(24)   \begin{align*}\left \langle x ,y ,z\right \rangle =\left \langle 1 ,2 ,3\right \rangle +t_{1}\left \langle -1 ,1 , -2\right \rangle \\ \left \langle x ,y ,z\right \rangle =\left \langle -1 ,0 ,2\right \rangle +t_{2}\left \langle 1 ,\frac{1}{2} , -2\right \rangle \end{align*}

Of course the two lines are not parallel since there is no k such that

(25)   \begin{equation*}\left \langle -1 ,1 , -2\right \rangle =k\left \langle 1 ,\frac{1}{2} , -2\right \rangle \end{equation*}

As we have seen before, the two vectors \left \langle -1 ,1 , -2\right \rangle and \left \langle 1 ,\frac{1}{2} , -2\right \rangle determine the direction of each line. Note that for the two vectors to be parallel to each other, we must have

(26)   \begin{align*}\left ( -1\right )k =1 \\ \left (1\right )k =\frac{1}{2} \\ \left ( -2\right )k = -2\end{align*}

The first equation says, k = -1 while the last equation claims k =1. If you believe, -1 \neq 1 then these two vectors are not parallel to each other. Now, do we know if the lines intersect each other? Let us find out. If the lines intersect, then there is a point that satisfies both equations. Assume the two lines intersect; define \left (x ,y ,z\right ) the point of intersection. Then from the first line

(27)   \begin{equation*}x =1 -t_{1} ,y =2 +t_{1} ,z =3 -2t_{1} \end{equation*}

and from the second line

(28)   \begin{equation*}x = -1 +t_{2} ,y =\frac{1}{2}t_{2} ,z =2 -2t_{2} \end{equation*}

Since \left (x ,y ,z\right ) is on both lines

(29)   \begin{align*}1 -t_{1} = -1 +t_{2} \\ 2 +t_{1} =\frac{1}{2}t_{2} \\ 3 -2t_{1} =2 -2t_{2}\end{align*}

We have three equations and two unknowns. From the second equation, we get

(30)   \begin{equation*}t_{2} =4 +2t_{1} \end{equation*}

Put this into the first equation

(31)   \begin{align*}1 -t_{1} = -1 +4 +2t_{1} \\ 1 +1 -4 =3t_{1} \\ t_{1} = -\frac{2}{3}\end{align*}

This means t_{2} =4 +2\left ( -\frac{2}{3}\right ) =\frac{8}{3}. If these values for t_{1} and t_{2} satisfy the last equation, we have found parameters which give us the point \left (x ,y ,z\right ); otherwise, there are no such parameters and in result no such point \left (x ,y ,z\right ) . The last equation dictates

(32)   \begin{align*}3 -2t_{1} =2 -2t_{2} \\ 3 -2( -\frac{2}{3}) =2 -2\genfrac{(}{)}{}{}{8}{3} \\ \frac{7}{3} = -\frac{16}{3}\end{align*}

This is not true, so our original assumption is not true. The two lines are not parallel and don’t intersect.

Example 23.

Suppose we are given a line \langle x ,y ,z\rangle=\left \langle 1 ,2 ,3\right \rangle +t\left \langle -1 ,0 ,2\right \rangle and a point \left (1 ,0 ,1\right ) . We would like to find the distance between these two. Let us list the given information first followed by a strategy.

1. The vector \left \langle -1 ,0 ,2\right \rangle can be used as a direction vector for the line.

2. The point \left (1 ,2 ,3\right ) is on the line.

3. Using the given points \left (1 ,0 ,1\right ) and \left (1 ,2 ,3\right ) we can construct a vector, say \left \langle 1 -1 ,0 -2 ,1 -3\right \rangle =\left \langle 0, -2 , -2\right \rangle.

4. Projection vector of \left \langle 0 , -2 , -2\right \rangle onto the vector on the line \left \langle -1 ,0 ,2\right \rangle is computed by

(33)   \begin{equation*}\frac{\left \langle 0 , -2 , -2\right \rangle \cdot \left \langle -1 ,0 ,2\right \rangle }{\left \langle -1 ,0 ,2\right \rangle \cdot \left \langle -1 ,0 ,2\right \rangle }\left \langle -1 ,0 ,2\right \rangle =\frac{ -4}{5}\left \langle -1 ,0 ,2\right \rangle =\left \langle \frac{4}{5} ,0 , -\frac{8}{5}\right \rangle \end{equation*}

5. We have many options. One way is that if we go along the line from the point \left (1 ,2 ,3\right ) by the projection vector that we just constructed, we find the closest point on the line to the point outside of the line \left (1 ,0 ,1\right ) . This way, we can find the distance between the two points and we are done.

(34)   \begin{equation*}\left \langle 1 ,2 ,3\right \rangle +\left \langle \frac{4}{5} ,0 , -\frac{8}{5}\right \rangle =\left \langle \frac{9}{5} ,2 ,\frac{7}{5}\right \rangle \end{equation*}

Now the distance between \left (1 ,0 ,1\right ) and \left (\frac{9}{5} ,2 ,\frac{7}{5}\right ) is

(35)   \begin{equation*}\sqrt{\left (1 -\frac{9}{5}\right )^{2} +(0 -2)^{2} +(1 -\frac{7}{5})^{2}} \end{equation*}

And we are done.

Example 24.

Suppose we are given a plane x +2y +3z =1 and the point \left (1 ,1 ,2\right ). We would like to find the distance between the point and the given plane. Let us again list the information and a strategy:

1. The point \left (1 ,0 ,0\right ) is on the plane.

2. The vector \left \langle 1 ,2 ,3\right \rangle is normal to the plane.

3. We can use a point on the plane and a point outside to make a vector, \left \langle 1 -1 ,1 -0 ,1 -0\right \rangle =\left \langle 0 ,1 ,1\right \rangle.

4. If we project this vector onto the normal vector to the plane, we can find its length and done.

(36)   \begin{equation*}proj_{\left \langle 1 ,2 ,3\right \rangle }\left \langle 0 ,1 ,1\right \rangle =\frac{\left \langle 1 ,2 ,3\right \rangle \cdot \left \langle 0 ,1 ,1\right \rangle }{\left \langle 0 ,1 ,1\right \rangle \cdot \left \langle 0 ,1 ,1\right \rangle }\left \langle 0 ,1 ,1\right \rangle =\frac{5}{2}\left \langle 0 ,1 ,1\right \rangle \end{equation*}

and magnitude of this vector is

(37)   \begin{equation*}\sqrt{0^{2} +(\frac{5}{2})^{2} +(\frac{5}{2})^{2}} \end{equation*}

In the event that we have to find the distance between two planes, the process is as easy (just choose a point on a plane and find the distance from this point to the other plane.)

Example 25.

Suppose we are given two planes x +2y +3z =1 and y =z +x . We know if the two planes are not parallel to each other, then they intersect in a line. We are interested to find a parametric equation for such line. Again, there are many ways to do this. For this exercise, we compare the normals to the planes \left \langle 1 ,2 ,3\right \rangle and \left \langle -1 ,1 , -1\right \rangle to see that the two planes are not parallel to each other. To construct an equation of any line in space, we need a point on the line and a direction vector. Finding a point is easy; put x =0 to get y =\frac{1}{5} =z . So the point \left (0 ,\frac{1}{5} ,\frac{1}{5}\right ) is on both planes. To find a direction vector for the line, observe that cross product of the vectors normal to each plane provides a new vector normal to both vectors. With the same token, this new vector is in the direction of the intersection line. So

(38)   \begin{align*}\left \langle 1 ,2 ,3\right \rangle \times \left \langle -1 ,1 , -1\right \rangle =\left ( -2 -3\right )\boldsymbol{i} -\left ( -1 -\left ( -3\right )\right )\boldsymbol{j} +\left (1 -\left ( -2\right )\right )\boldsymbol{k} \\ = -5\boldsymbol{i} -2\boldsymbol{j} +3\boldsymbol{k}\end{align*}

The line

(39)   \begin{equation*}x =0 -5t ,y =\frac{1}{5} -2t ,z =\frac{1}{5} +3t \end{equation*}

is the line of intersection.

Please see a few additional problems for this section.

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