Additional Vector Examples
Additional Substitutions

Example 1.

We start by evaluating a cross product

(1)   \begin{align*}\langle 2 ,1 , -7\rangle \times \langle -3 ,5 , -4\rangle =\det \left [\begin{array}{ccc}\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 2 & 1 & -7 \\ -3 & 5 & -4\end{array}\right ] \\ =( -4 +35)\boldsymbol{i} -( -8 -21)\boldsymbol{j} +(10 +3)\boldsymbol{k} \\ =\left \langle 31 ,29 ,13\right \rangle \end{align*}

Example 2.

Find the area of a parallelogram with side vectors \overrightarrow{PQ} and \overrightarrow{PR} where P\left ( -4 ,6 ,3\right ) ,Q\left (1 ,1 , -5\right ) ,R\left (2 ,2 ,2\right ). We start by finding the two desired vectors

(2)   \begin{equation*}\overrightarrow{PQ} =\left \langle 1 +4 ,1 -6 , -5 -3\right \rangle =\left \langle 5 , -5 , -8\right \rangle \end{equation*}

(3)   \begin{equation*}\overrightarrow{PR} =\left \langle 2 +4 ,2 -6 ,2 -3\right \rangle =\left \langle 6 , -4 , -1\right \rangle \end{equation*}

Now we the cross product of these two. Note that, in this case order does not matter since we will find the magnitude after.

(4)   \begin{align*}\det \left [\begin{array}{ccc}\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 5 & -5 & -8 \\ 6 & -4 & -1\end{array}\right ] =\left (5 -32\right )\boldsymbol{i} -\left ( -5 +48\right )\boldsymbol{j} +\left ( -20 +30\right )\boldsymbol{k} \\ =\left \langle -27 , -43 ,10\right \rangle \end{align*}

And area of the given parallelogram is equal to

(5)   \begin{equation*}\left \Vert \left \langle -27 , -43 ,10\right \rangle \right \Vert =\sqrt{\left ( -27\right )^{2} +( -43)^{2} +(10)^{2}} \end{equation*}

Example 3.

Suppose we are given two lines

(6)   \begin{align*}\frac{x}{3} =\frac{y -2}{4} =\frac{z +4}{1} \\ \frac{x +6}{3} =\frac{y +2}{4} =\frac{z -3}{2}\end{align*}

We know the two lines are not parallel due to the fact that the direction vectors of these two lines \left \langle 3 ,4 ,1\right \rangle and \left \langle 3 ,4 ,2\right \rangle are not parallel to each other. If you believe otherwise, try to find a constant a such that one vector is a constant multiple of the other. Now let us see if the two lines intersect. There are many ways to do this. Let us first write the parametric form of each line in order

(7)   \begin{align*}x =3t ,y =4t +2 ,z =t -4 \\ x =3r -6 ,y =4r -2 ,z =2r +3\end{align*}

Now if \left (x ,y ,z\right ) is a point of intersection, for this point we must have

(8)   \begin{align*}3t =3r -6 \\ 4t +2 =4r -2 \\ t -4 =2r +3\end{align*}

Solving this system we get

(9)   \begin{align*}t =r -2 \\ 4\left (r -2\right ) +2 =4r -2\end{align*}

The second equation claims that 6 =2 which is not true. So the two lines do not intersect with each other.

Example 4.

We would like to find the distance between the two parallel planes x -2z = -1 and x -2z =4. Let us choose a point on each plane, in order \left ( -1 ,0 ,0\right ) and \left (4 ,0 ,0\right ) . Next construct a vector using this two points \left \langle 5 ,0 ,0\right \rangle . We can now project this vector onto the normal (to any of the two planes.)

(10)   \begin{align*}\text{proj}_{\left \langle 1 ,0 , -2\right \rangle }\left \langle 5 ,0 ,0\right \rangle =\frac{\left \langle 5 ,0 ,0\right \rangle \cdot \left \langle 1 ,0 , -2\right \rangle }{\left \Vert \left \langle 5 ,0 ,0\right \rangle \right \Vert ^{2}}\left \langle 5 ,0 ,0\right \rangle \\ =\frac{5}{25}\left \langle 5 ,0 ,0\right \rangle =\boldsymbol{i}\end{align*}

Now length of this vector provides the distance between the two planes that is \left \Vert \boldsymbol{i}\right \Vert =1.

Example 5.

We want to find equation of the intersection line of two planes x -y =3 and y -2x =2. Note that to write equation of any line, we need a point on the line and a direction vector along the line. To find a point, we try to solve the following system. Remember that the coordinates that we find for x and y can have any z value in this case.

(11)   \begin{align*}x -y =3 \\ y -2x =2\end{align*}

This means

(12)   \begin{equation*}x = -5 ,y = -8 \end{equation*}

Take z =0 and we are good to go. Now that we chose the point \left ( -5 , -8 ,0\right ), we need a direction vector along the line intersection of two planes. As we discussed before, the cross product of two normal lines to each plane will provide a vector that is perpendicular to both normals and therefore in direction of the desired line. So

(13)   \begin{align*}\left \langle 1 , -1 ,0\right \rangle \times \left \langle -2 ,1 ,0\right \rangle =\det \left [\begin{array}{ccc}\boldsymbol{i} &\boldsymbol{j} &\boldsymbol{k} \\ 1 & -1 & 0 \\ -2 & 1 & 0\end{array}\right ] \\ =\left \langle 0 ,0 , -1\right \rangle \end{align*}

So the equation of the desired line is

(14)   \begin{equation*}\left \langle x ,y ,z\right \rangle =\left \langle -5 , -8 ,0\right \rangle +t\left \langle 0 ,0 , -1\right \rangle \end{equation*}

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