Vectors. Equations of Lines and Planes
Introduction
Vectors are main objects of study in multivariable calculus. They have the same role that numbers have in single-variable calculus. It is very important to gain a solid understanding of vectors before proceeding to multivariable calculus. Please read this section carefully, and make sure you are able to solve most of the exercises in quizzes in the bottom of this page.
Vectors: Geometric point of view
Before we can see why vectors may be useful in calculus, let us review our former knowledge about main properties of vectors.
Vector: A wrong definition
Vector is a line segment with an arrow on one of its ends.
If you don’t like the previous definition, you are right, and you should be proud. But we won’t provide a better one in this page. This website does have (or will have) a precise definition of the vector in the appropriate section. Here we will just review how vectors work.
Equal vectors
Two vectors are equal if they lie on parallel lines, have the same direction, and have the same magnitude.
In the picture above, we have \(\overrightarrow{AB}=\overrightarrow{CD}\), while \(\overrightarrow{AB}\neq \overrightarrow{EF}\).
You may have noticed that the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{EF}\) have the same length. That’s right.
We first made \(\overrightarrow{AB}\), then we copied it and rotated to obtain \(\overrightarrow{EF}\). The vectors \(\overrightarrow{AB}\) and \(\overrightarrow{EF}\) have the same length, but they are different vectors.
We write \(|\overrightarrow{AB}|=|\overrightarrow{EF}|\) to say that the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{EF}\) have the same length.
Vector addition
Adding two vectors is a geometric operation that produces a
vector.
If one vector starts exactly where the other ends, then the addition is easy. This case is illustrated by the picture above.
We have \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\).
In the example from the picture above, we want to find \(\overrightarrow{AB}+\overrightarrow{CD}\). These two vectors are drawn in black. We will reduce this case to the previous one in which one vector starts where the other ends. We first construct a point \(E\) such that \(\overrightarrow{CD}=\overrightarrow{BE}\). Then we have
\[\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AE}.\]
The point \(E\) is blue in the picture, as well as the vector \(\overrightarrow{BE}\). The resulting vector \(\overrightarrow{AE}\) is also blue.
The opposite vector to the vector \(\overrightarrow{AB}\) is the vector \(\overrightarrow{BA}\). We also write
\[-\overrightarrow{AB}=\overrightarrow{BA}.\]
Now we can define subtraction of vectors: \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AB}+(-\overrightarrow{CD})=\overrightarrow{AB}+\overrightarrow{DC}\).
Multiplying vector by a scalar
The vector \(2\overrightarrow{AB}\) is the vector that is parallel to \(\overrightarrow{AB}\), has the same direction, but is twice the length of \(\overrightarrow{AB}\). Similarly, we define \(3\overrightarrow{AB}\), and \(\pi\overrightarrow{AB}\).
Vectors: Algebraic point of view
This is the beginning of the multivariable calculus. We will start by encoding vectors in a plane using two variables.
Two dimensions
When we say two dimensions, we mean that we are dealing with a plane. Let us choose one point in the plane and denote it by
O. This point will be called the origin. Let us place two vectors of length 1 that start at
O and are perpendicular to each other. Let us call them \(\overrightarrow i\) and \(\overrightarrow j\). Then for each point \(A\) in the plane we can express \(\overrightarrow{OA}\) using only vectors \(i\) and \(j\).
In the previous picture we can represent \(\overrightarrow{OB}=\overrightarrow{i}+2\overrightarrow{j}\), and \(\overrightarrow{OA}=-\overrightarrow{i}+1.5\overrightarrow{j}\). We also say that each vector is a linear combination of \(\overrightarrow i\) and \(\overrightarrow j\).
Each vector in the plane can be written in terms of \(\overrightarrow{i}\) and \(\overrightarrow{j}\). We can write
\[\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{i}-1.5\overrightarrow{j}+\overrightarrow{i}+2\overrightarrow{j}=2\overrightarrow i+0.5\overrightarrow j.\]
We say that 2 and 0.5 are
coordinates of the vector \(\overrightarrow{AB}\) and instead of \(\overrightarrow{AB}=2\overrightarrow i+0.5\overrightarrow j\) we will often write \(\overrightarrow{AB}=\langle 2,0.5\rangle\). Similarly, \(\overrightarrow{OA}=\langle -1,1.5\rangle\) and \(\overrightarrow{OB}=\langle 1,2\rangle\).
Vectors in three dimensions
The story is exactly the same. We fix one point,
O and three unit vectors \(\overrightarrow i\), \(\overrightarrow j\), and \(\overrightarrow k\). We can express each vectors as a linear combination of \(\overrightarrow i\), \(\overrightarrow j\), and \(\overrightarrow k\). We also write vectors in coordinate notation, so that \(\langle 3, 5, -2\rangle\) represents the same vector as \(3\overrightarrow i+5\overrightarrow j-2\overrightarrow k\).
Dot product of vectors
If two vectors are written in their coordinate notation, \(\overrightarrow u=\langle u_x,u_y,u_z\rangle\) and \(\overrightarrow v=\langle v_x, v_y,v_z\rangle\) we define the
scalar product (often called the
dot product) of vectors \(\overrightarrow u\) and \(\overrightarrow v\) in the following way:
\[\overrightarrow u\cdot\overrightarrow v=u_xv_x+u_yv_y+u_zv_z.\]
The dot product of two vectors is always a scalar. For example, \(\langle 3,2,-1\rangle\cdot \langle -4,5,2\rangle=3\cdot (-4)+2\cdot 5+(-1)\cdot 2=-4\).
The following properties of the dot product are very easy to verify. However, they will have powerful consequences:
Theorem (Properties of dot product)
\[\overrightarrow u\cdot\overrightarrow v=\overrightarrow v\cdot\overrightarrow u;\]
\[\overrightarrow u\cdot \overrightarrow u=|\overrightarrow u|\geq 0;\]
\[\overrightarrow u\cdot (\alpha \overrightarrow v)=(\alpha\overrightarrow u)\cdot \overrightarrow v=\alpha (\overrightarrow u\cdot\overrightarrow v);\]
\[\overrightarrow u\cdot(\overrightarrow v+\overrightarrow w)=\overrightarrow u\cdot\overrightarrow v+\overrightarrow u\cdot\overrightarrow w.\]
Theorem (Cosine theorem)
Denote by \(\theta\) the angle between vector \(\overrightarrow u\) and \(\overrightarrow v\) (measured once the vectors are placed to have the same starting point). Then we have
\[\overrightarrow u\cdot \overrightarrow v=|\overrightarrow u|\cdot|\overrightarrow v|\cos\theta.\]
Let us place the vectors in the plane such that \(\overrightarrow{AB}=\overrightarrow u\), \(\overrightarrow{AC}=\overrightarrow v\). Then \(\theta=\angle BAC\).
According to the cosine theorem from geometry we have:
\[\left|\overrightarrow{BC}\right|^2=\left|\overrightarrow{AC}\right|^2+\left|\overrightarrow{AB}\right|^2-2\left|\overrightarrow {AC}\right|\cdot\left|\overrightarrow{AB}\right|\cdot\cos\theta.\]
We have that \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow {AB}=\overrightarrow v-\overrightarrow u\), and using the properties of the dot product we have
\[|\overrightarrow{BC}|^2=\overrightarrow{BC}\cdot\overrightarrow{BC}=(\overrightarrow v-\overrightarrow u)^2=\overrightarrow v\cdot \overrightarrow v+\overrightarrow u\cdot \overrightarrow u-2\overrightarrow u\cdot\overrightarrow v.\]
This and the previous formula directly imply the desired result.
Cross product
We will start by formal definition of the cross product.
Definition (Cross product)
If \(\overrightarrow u=\langle x_u,y_u,v_u\rangle\) and \(\overrightarrow v=\langle x_v,y_v,z_v\rangle\)
are two vectors, then their cross product \(\overrightarrow u\times \overrightarrow v\) is defined as the following vector: \[\overrightarrow u\times\overrightarrow v=\langle y_uz_v-y_vz_u,-x_uz_v+x_vz_u,x_uy_v-x_vy_u\rangle.\]
It is easier to remember the cross product if it is written in the form of a determinant:
\[\overrightarrow u\times\overrightarrow v=\mbox{det }\left|\begin{array}{ccc} \overrightarrow i&\overrightarrow j&\overrightarrow k\newline x_u&y_u&z_u\newline x_v&y_v&z_v\end{array}\right|.\]
The following theorem lists the properties of the cross product. It is very easy to prove (using just the definition of the cross product, and following the order in which the statements are listed).
Theorem (Properties of cross product)
Assume that \(\overrightarrow u\), \(\overrightarrow v\), and \(\overrightarrow w\) are three vectors, and \(\alpha\), \(\beta\), and \(\gamma\) are three scalars.
Then:
- \(\overrightarrow u\times (\beta\overrightarrow v+\gamma\overrightarrow w)=\beta\overrightarrow u\times\overrightarrow v+\gamma\overrightarrow u\times\overrightarrow w\).
- \(\overrightarrow u\times\overrightarrow v=-\overrightarrow v\times\overrightarrow u\)
- \(\left|\overrightarrow u\times\overrightarrow v\right|^2+\left(\overrightarrow u\cdot\overrightarrow v\right)^2=\left|\overrightarrow u\right|^2\cdot\left|\overrightarrow v\right|^2\)
- \(\left|\overrightarrow u\times\overrightarrow v\right|=\left|\overrightarrow u\right|\cdot \left|\overrightarrow v\right|\cdot \left|\sin \theta\right|\), where \(\theta\) is the angle between vectors \(\overrightarrow u\) and \(\overrightarrow v\). This implies that the length of the cross product is equal to the area of the parallelogram determined by \(\overrightarrow u\) and \(\overrightarrow v\).
-
The vector \(\overrightarrow u\times \overrightarrow v\) is perpendicular to both \(\overrightarrow u\) and \(\overrightarrow v\).
Triple product
Definition (Triple product)
If \(\overrightarrow u\), \(\overrightarrow v\), and \(\overrightarrow w\) are three vectors, then we define their triple product as \[\left[\overrightarrow u, \overrightarrow v, \overrightarrow w\right]=\overrightarrow u\cdot \left(\overrightarrow v\times\overrightarrow w\right).\]
Notice that the result of a triple product is a scalar, and the absolute value of that scalar is equal to the volume of the parallelepiped spanned by the vectors \(\overrightarrow u\), \(\overrightarrow v\), and \(\overrightarrow w\).
Equations of lines
You are probably familiar with the equations of the form \(y=3x+5\). There are many ordered pairs \((x,y)\) of real numbers. Some of them satisfy the previous equation, others don’t. Those that do satisfy the equation determine the points in the plane - and if you paint those points in green - you get a green line.
This is the basic principle behind equations of geometric objects. A line in geometry is uniquely determined by a point and a vector. Let us consider a line \(\ell\) that is determined by a point \(A(x_0,y_0,z_0)\) and a vector \(v=\langle a,b,c\rangle\). We want to describe all those points \(P(x,y,z)\) that belong to \(l\). Geometric description is easy:
The point \(P(x,y,z)\) belongs to \(\ell\) if and only if \(\overrightarrow {AP}\) is parallel to \(\overrightarrow v\).
The last sentence has alternative form:
The point \(P(x,y,z)\) belongs to the line \(\ell\) if and only if there exists a real number \(t\) such that \(\overrightarrow {AP}=t\cdot \overrightarrow v\).
Let us denote the origin by \(O\). Notice that \[\overrightarrow{AP}=\overrightarrow{AO}+\overrightarrow{OP}=\overrightarrow{OP}-\overrightarrow{OA}=\langle x,y,z\rangle-\langle x_0,y_0,z_0\rangle=\langle x-x_0,y-y_0,z-z_0\rangle.\] The last sentence can now be easily translated into algebraic form:
The point \(P(x,y,z)\) belongs to the line \(\ell\) if and only if there exists a real number \(t\) such that \(\langle x-x_0,y-y_0,z-z_0\rangle=t\langle a,b,c\rangle\).
The last form of the equation of the line is called the vector equation of the line \(\ell\). Its equivalent version is used most often and is called the parametric equation of the line:
\[x=x_0+at\]
\[y=y_0+bt\]
\[z=z_0+bt\]
\[t\in\mathbb R.\]
Equations of planes
A plane in three dimensional space is uniquely determined by one of the following three combinations :
- (i) any three of its points
- (ii) a point and any two vectors
- (iii)a point and a normal vector
It turns out that in practice we use (ii) and (iii) to determine the equations of the plane. If we are given three points, we transform the problem from (i) to either (ii) or (iii) and try to deal with consequences.
Case (ii): Parametric equations
Let us assume that we are given a point \(A\) and two vectors \(\overrightarrow u=\langle a_u,b_u,c_u\rangle\) and \(\overrightarrow v=\langle a_v,b_v,c_v\rangle\). Let us consider the plane \(\alpha\) that passes through \(A\) and is parallel to both \(\overrightarrow u\) and \(\overrightarrow v\). A points \(P(x,y,z)\) belongs to \(\alpha\) if and only if there are two real numbers \(s\) and \(t\) such that \(\overrightarrow{AP}= s\overrightarrow u+t\overrightarrow v\), i.e. \(\overrightarrow{OP}=\overrightarrow{OA}+s\overrightarrow u+t\overrightarrow v\). This can be written as \[x=x_0+sa_u+ta_v\] \[y=y_0+sb_u+tb_v\] \[z=z_0+sc_u+tc_v\] \[s\in\mathbb R, \;\;\; t\in\mathbb R.\]
The above equations are called parametric equations of the plane \(\alpha\).
Notice that the parametric equations of the line have one parameter, while the parametric equations of the plane have two parameters (\(s\) and \(t\)). That is why people say that line has dimension 1, while plane has dimension 2.
Case (iii): Symmetric equations
Assume we are given a point \(A(x_0,y_0,z_0)\) and a vector \(\overrightarrow n=\langle a,b,c\rangle\). Let us find the equation of the plane \(\alpha\) that passes through \(A\) and is orthogonal to \(\overrightarrow n\). A point \(P(x,y,z)\) belongs to \(\alpha\) if and only if the vector \(\overrightarrow{AP}\) is orthogonal to \(\overrightarrow n\), i.e. if and only if \(\overrightarrow{AP}\cdot \overrightarrow n=0\). Since \(\overrightarrow{AP}=\langle x-x_0, y-y_0, z-z_0\rangle\), the last statment can be written as \[(x,y,z)\in\alpha\;\;\;\Leftrightarrow \;\;\; \langle x-x_0,y-y_0,z-z_0\rangle \cdot \langle a,b,c\rangle=0,\;\;\;\mbox{ or, equivalently}\]
\[a(x-x_0)+b(y-y_0)+c(z-z_0)=0.\]
Sometimes the last equation is written as \(ax+by+cz=d\), where \(d=ax_0+by_0+cz_0\) is a constant.
Theorem (Distance between a point and a plane)
Assume that \(A(x_1,y_1,z_1)\) is a point, and \(\alpha\) is a plane given by \[a(x-x_0)+b(y-y_0)+c(z-z_0)=0.\]
Then the distance between the point \(A\) and the plane \(\alpha\) is given by \[\mbox{dist }(A,\alpha)=\frac{\left|a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)\right|}{\sqrt{a^2+b^2+c^2}}.\]
Let \(X\in\alpha\) be the point such that \(AX\perp \alpha\). Then \(\mbox{dist }(A,\alpha)=\left|\overrightarrow{AX}\right|\). Notice that \(\overrightarrow n=\langle a,b,c\rangle\) is a normal vector to \(\alpha\). Now we need to find a point \(X(x^{\prime},y^{\prime},z^{\prime})\) such that \(\overrightarrow {AX}\|\overrightarrow n\). We must have \(\overrightarrow {AX}=\langle x^{\prime}-x_1, y^{\prime}-y_1, z^{\prime}-z_1\rangle =\lambda \langle a,b,c\rangle \) for some \(\lambda \in\mathbb R\). Therefore, we are looking for \(\lambda \in \mathbb R\) such that the point \((x^{\prime},y^{\prime},z^{\prime})=(x_1+\lambda a, y_1+\lambda b, z_1+\lambda c)\) belongs to \(\alpha\). This is equivalent to finding a real number \(\lambda\) such that \[a(x_1+\lambda a-x_0)+
b(y_1+\lambda b-y_0)+c(z_1+\lambda c-z_0)=0\]
\[ \Leftrightarrow \;\;\; \lambda=-\frac{a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)}{a^2+b^2+c^2}.\]
We now have \[\left|\overrightarrow{AX}\right|=\left|\langle x^{\prime}-x_1,y^{\prime}-y_1,z^{\prime}-z_1\rangle\right|
=\left|\lambda\right|\cdot \left|\langle a,b,c\rangle\right|=\frac{\left|a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)\right|}{\sqrt{a^2+b^2+c^2}}.\]
Example
A plane is given by its symmetric equation \(8x-7y+5z=21\). Find the point on the plane that is closest to the point \(P\) with coordinates \((3,1,-2)\).
Let us denote by \(\alpha\) the given plane.
The required point is the intersection of \(\alpha\) with the line through \(P\) orthogonal to \(\alpha\). The normal vector to \(\alpha\) is \(\langle 8,-7,5\rangle\). The equation of the line through \(P\) orthogonal to \(\alpha\) is:
\begin{eqnarray*}
x&=&3+ 8t\\
y&=& 1-7t\\
z&=& -2+5t.
\end{eqnarray*}
The intersetion of the line with \(\alpha\) is the point \((3+8t, 1-7t, -2+5t)\) for which
\(8(3+8t)-7(1-7t)+5(-2+5t)=21\). Solving for \(t\) yields: \(24-7-10+(64+49+25)t=21\), which implies that \(t=\frac{14}{138}=\frac{7}{69}\). Thus the required point is \(Q\left(3+\frac{56}{69}, 1-\frac{49}{69}, -2+\frac{35}{69}\right)\).
Practice problems
Problem 1.
If \(A=(0,3,1)\), \(B=(2,1,1)\), and \(C=(1,4,-1)\) find the dot product \(\overrightarrow{AB}\cdot\overrightarrow{CB}\).
The coordinates of the point \(A\) are \((0,3,1)\) which means that \(\overrightarrow{OA}=\langle 0,3,1\rangle\), where \(O\) is the origin. Similar identities hold for \(\overrightarrow{OB}\) and \(\overrightarrow{OC}\). Now we can find the vector \(\overrightarrow{AB}\) in the following way: \[\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{OB}-\overrightarrow{OA}=\langle 2,1,1\rangle-\langle 0,3,1\rangle=\langle 2,-2,0\rangle.\]
Similarly, \[\overrightarrow{CB}=\overrightarrow{CO}+\overrightarrow{OB}=\overrightarrow{OB}-\overrightarrow{OC}=\langle 2,1,1\rangle-\langle 1,4,-1\rangle=\langle 1,-3,2\rangle.\]
Now the dot product can be found as \[\overrightarrow{AB}\cdot\overrightarrow{CB}=\langle 2,-2,0\rangle\cdot \langle 1,-3,2\rangle=2+6+0=8.\]
Problem 2.
Find the distance between the point \(A(3,1,0)\) and the plane determined by the points: \(B(0,1,2)\), \(C(-1,3,1)\), and \(D(3,-2,-1)\).
We need the symmetric equation of the plane \(\alpha\) determined by the points \(B\), \(C\), and \(D\). In order to obtain it, we first find a normal vector. Cross product of any two vectors in the plane give a normal vector. In order to find vectors, we use the origin \(O(0,0,0)\) to help us transform the coordinates of points into coordinates of vectors. \[\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\langle -1,3,1\rangle-\langle 0,1,2\rangle=\langle -1,2,-1\rangle.\]
Similarly, \[\overrightarrow{BD}=\overrightarrow{OD}-\overrightarrow{OB}=\langle 3,-2,-1\rangle-\langle 0,1,2\rangle=\langle 3, -3, -3\rangle.\]
Therefore \[\overrightarrow{BC}\times\overrightarrow{BD}=\mbox{det }\left| \begin{array}{ccc}\overrightarrow i&\overrightarrow j&\overrightarrow k\newline -1 & 2 &-1\newline 3& -3 & -3\end{array}\right|=-10\overrightarrow i-6\overrightarrow j-3\overrightarrow k=\langle -10,-6,-3\rangle.\]
Since we have a normal vector, the equation of the plane is \[-10\cdot (x-x_0)-6\cdot (y-y_0)-3\cdot (z-z_0)=0.\quad\quad\quad\quad\quad (\star)\]
Here \((x_0,y_0,z_0)\) is a fixed point in a plane. \((x,y,z)\) are coordinates of a variable point. Some points do satisfy \((\star)\) and some don’t. Those that do - they belong to the plane, and that is why \((\star)\) is called the equation of the plane.
The equation of our plane is obtained when we replace \((x_0,y_0,z_0)\) with coordinates of any point that belongs to the plane. By picking the point \(B\) we obtain: \[-10(x-0)-6\cdot (y-1)-3\cdot (z-2)=0.\] Using the formula for the distance from a point to the plane we now obtain:
\[\mbox{dist }(A, \alpha)=\frac{\left|-10\cdot (3-0)-6\cdot (1-1)-3\cdot (0-2)\right|}{\sqrt{(-10)^2+(-6)^2+(-3)^2}}=\frac{24}{\sqrt{173}}.
\]
Problem 3.
Assume that \(A(1,5,2)\), \(B(-3,4,-1)\), and \(C(0,-2,1)\) are vertices of the triangle. Let \(M\) be the midpoint of the segment \(AB\). Determine \(\overrightarrow{AB}\cdot \overrightarrow{CM}\).
We need to determine the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{CM}\).
It is rather straightforward to find \(\overrightarrow{AB}\) by expressing this vector as a sum of vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\). Here \(O\) is the origin and the coordinate definitions of \(A\) and \(B\) give us \(\overrightarrow{OA}=\langle 1,5,2\rangle\) and \(\overrightarrow{OB}=\langle -3,4,-1\rangle\). Therefore \[\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{OB}-\overrightarrow{OA}=\langle -3,4,-1\rangle-\langle 1,5,2\rangle=\langle -4,-1,-3\rangle.\]
We need to determine the vector \(\overrightarrow{CM}\), and this can be done by expressing \(\overrightarrow{CM}=\overrightarrow{OC}-\overrightarrow{OM}\). For this to work, we need to find the vector \(\overrightarrow{OM}\). We will now investigate the point \(M\) further.
So far, all we know about \(M\) is that it is the midpoint of \(AB\). This information can be written in the language of vectors: \(\overrightarrow{AM}=\overrightarrow{MB}\). Since \(\overrightarrow{AM}+\overrightarrow{MB}=\overrightarrow{AB}\) we get \(\overrightarrow{AM}=\frac12\overrightarrow{AB}\). We have \(\overrightarrow{AB}=\langle -4,-1,-3\rangle\), hence \(\overrightarrow{AM}=\left\langle -2,\frac{-1}2,\frac{-3}2\right\rangle\). We now have \[\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\langle 1,5,2\rangle+\left\langle -2,\frac{-1}2,\frac{-3}2\right\rangle=\left\langle -1,\frac{9}2,\frac{1}2\right\rangle.\]
Therefore \[\overrightarrow{CM}=\overrightarrow{OM}-\overrightarrow{OC}=\left\langle -1,\frac92,\frac12\right\rangle-\langle0,-2,1\rangle=\left\langle -1,\frac{13}2, -\frac12\right\rangle.\]
We can now find the required dot product:
\[\overrightarrow{AB}\cdot\overrightarrow{CM}=\langle-4,-1,-3\rangle\cdot \left\langle -1,\frac{13}2,-\frac12\right\rangle= 4-\frac{13}2+\frac{3}2=\frac{8-13+3}2=-1.\]
Problem 4.
Consider the following six points:
\[A(3,1,-2),\;\;\; B(-2,1,-5),\;\;\; C(3,0,4),\]
\[D(1,0,-1),\;\;\; E(-4,0,1),\;\;\; F(6,0,7).\]
Determine which of the following statements is true.
- (A) All six points belong to a plane.
- (B) None four of the given points belong to a plane.
- (C) The points \(A\), \(B\), \(E\), and \(F\) belong to a plane.
- (D) The points \(A\), \(B\), \(D\), and \(E\) belong to a plane.
- (E) The points \(A\), \(B\), \(C\), and \(F\) belong to a plane.
The points \(A\), \(B\), \(E\), and \(F\) belong to a plane if and only if \(\left[\overrightarrow{AB},\overrightarrow{AE},\overrightarrow{AF}\right]=0\). This follows from the fact that \(\left|\left[\overrightarrow{AB},\overrightarrow{AE},\overrightarrow{AF}\right]\right|\) represents the volume of the parallelepiped spanned by these three vectors. In order to determine the vectors \(\overrightarrow{AB}\), \(\overrightarrow{AE}\), \(\overrightarrow{AF}\) we denote by \(O\) the origin and notice that \(\overrightarrow{OA}=\langle 3,1,-2\rangle\), \(\overrightarrow{OB}=\langle -2,1,-5\rangle\), etc. Therefore
\[\overrightarrow{AB}= \overrightarrow{OB}-\overrightarrow{OA}=\langle-2,1,-5\rangle-\langle3,1,-2\rangle=\langle -5,0,-3\rangle;\]
\[\overrightarrow{AE}= \overrightarrow{OE}-\overrightarrow{OA}=\langle-4,0,1\rangle-\langle3,1,-2\rangle=\langle -7,-1,3\rangle;\]
\[\overrightarrow{AF}= \overrightarrow{OE}-\overrightarrow{OA}=\langle6,0,7\rangle-\langle3,1,-2\rangle=\langle 3,-1,9\rangle.\]
We now have:
\[\left[\overrightarrow{AB},\overrightarrow{AE},\overrightarrow{AF}\right]=\mbox{det }\left|
\begin{array}{ccc} -5&0&-3\newline -7&-1&3\newline 3&-1&9\end{array}\right|=(-5)\cdot \mbox{det }\left|\begin{array}{cc} -1&3\newline -1&9\end{array}\right| -0\cdot \mbox{det }\left|\begin{array}{cc} -7&3\newline 3&9\end{array}\right|+(-3)\cdot \mbox{det }\left|\begin{array}{cc}-7 & -1\newline 3&-1\end{array}
\right|.\]
This gives us: \[\left[\overrightarrow{AB},\overrightarrow{AE},\overrightarrow{AF}\right]=
-5\cdot (-9+3)-3\cdot (7+3)=-5\cdot (-6)-3\cdot 10=0,\] hence the points \(A\), \(B\), \(E\), and \(F\) belong to a plane.
Now we want to determine whether the points \(A\), \(B\), \(D\), and \(E\) belong to a plane. For this we need to find \(\left[\overrightarrow{AB}, \overrightarrow{AE}, \overrightarrow{AD}\right]\). The vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AE}\) are known from before and we only need to find \(\overrightarrow{AD}\):
\[\overrightarrow{AD}= \overrightarrow{OD}-\overrightarrow{OA}=\langle1,0,-1\rangle-\langle3,1,-2\rangle=\langle -2,-1,1\rangle.\]
We now have:
\[\left[\overrightarrow{AB},\overrightarrow{AE},\overrightarrow{AD}\right]=\mbox{det }\left|
\begin{array}{ccc} -5&0&-3\newline -7&-1&3\newline -2&-1&1\end{array}\right|=-5\cdot \mbox{det }\left|\begin{array}{cc} -1&3\newline -1&1\end{array}\right| -0\cdot \mbox{det }\left|\begin{array}{cc} -7&3\newline -2&1\end{array}\right|+(-3)\cdot \mbox{det }\left|\begin{array}{cc}-7 & -1\newline -2&-1\end{array}
\right|=-25.\]
Since \(-25\neq 0\), we have that the points \(A\), \(B\), \(D\), and \(E\) do not lie in a plane.
Our next task is to find out whether the points \(A\), \(B\), \(C\), and \(F\) belong to a plane. Since \(A\), \(B\), \(E\), and \(F\) belong to a plane, this is the same as checking whether \(C\) belongs to the plane determined by \(A\), \(B\), and \(E\). For this we need to find \(\left[\overrightarrow{AB}, \overrightarrow{AE}, \overrightarrow{AC}\right]\). The vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AE}\) are known from before and we only need to find \(\overrightarrow{AC}\):
\[\overrightarrow{AC}= \overrightarrow{OC}-\overrightarrow{OA}=\langle3,0,4\rangle-\langle3,1,-2\rangle=\langle 0,-1,6\rangle.\]
We now have:
\[\left[\overrightarrow{AB},\overrightarrow{AE},\overrightarrow{AC}\right]=\mbox{det }\left|
\begin{array}{ccc} -5&0&-3\newline -7&-1&3\newline 0&-1&6\end{array}\right|=-5\cdot \mbox{det }\left|\begin{array}{cc} -1&3\newline -1&6\end{array}\right| -0\cdot \mbox{det }\left|\begin{array}{cc} -7&3\newline 0&6\end{array}\right|+(-3)\cdot \mbox{det }\left|\begin{array}{cc}-7 & -1\newline 0&-1\end{array}
\right|=-6.\]
Since \(-6\neq 0\), we have that the points \(A\), \(B\), \(C\), and \(E\) do not lie in a plane.
Hence, C is the only correct answer.
Problem 5.
Consider the points \(O(0,0,0)\), \(A(3,-1,0)\), \(B(-1,2,-3)\), and \(C(-2,1,2)\). Find the shortest distance between the lines \(OA\) and \(BC\).
Since \[\left[\overrightarrow{OA}, \overrightarrow{OB},\overrightarrow{OC}\right]=\mbox{det }\left|\begin{array}{ccc}3&-1&0\newline-1&2&-3\newline-2&1&2\end{array}\right|\neq 0,\]
the vectors \(\overrightarrow{OA}\), \(\overrightarrow{OB}\), and \(\overrightarrow{OC}\) form a parallelepiped of a non-zero volume. This means that \(O\), \(A\), \(B\), \(C\) are not in a plane, and this further implies that the lines \(OA\) and \(BC\) are neither parallel nor intersecting.
There is the unique plane \(\alpha\) that contains \(OA\) and is parallel to \(BC\), and there is the unique plane \(\beta\) that contains \(BC\) and is parallel to \(OA\). The planes \(\alpha\) and \(\beta\) are parallel. The shortest distance between the lines \(OA\) and \(BC\) is equal to the distance between the planes \(\alpha\) and \(\beta\). The distance between the planes \(\alpha\) and \(\beta\) is equal to the distance between the point \(B\) and the plane \(\alpha\).
In order to find the distance between the point \(B\) and the plane \(\alpha\) we will use the formula
\[\mbox{dist }(B, \alpha)=\frac{\left|ax_0+by_0+cz_0-d\right|}{\sqrt{a^2+b^2+c^2}},\]
where \((x_0,y_0,z_0)\) are coordinates of the point \(B\) and \(a\), \(b\), \(c\), \(d\) are the coefficients from the symmetric equation \(ax+by+cz-d=0\) for the plane \(\alpha\).
Let us now find \(a\), \(b\), \(c\), and \(d\). The coefficients \(a\), \(b\), \(c\) determine the normal vector, and we may choose them so that \(\langle a,b,c\rangle=\overrightarrow{OA}\times\overrightarrow{BC}\). The vector \(\overrightarrow{BC}\) can be found as \(\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\langle-1,-1,5 \rangle\). We now have: \[\overrightarrow{OA}\times\overrightarrow{BC}=\left|\begin{array}{ccc} \overrightarrow{i} & \overrightarrow{j}&\overrightarrow{k}\newline
3&-1&0\newline -1&-1&5\end{array}
\right|=-5\overrightarrow i-15\overrightarrow j-4\overrightarrow k=\langle -5,-15,-4\rangle.\]
Now the equation of the plane \(\alpha\) is: \[-5\cdot (x-x_{\alpha})-15\cdot (y-y_{\alpha})-4\cdot (z-z_{\alpha})=0,\] where \((x_{\alpha},y_{\alpha},z_{\alpha})\) is an arbitrary point of the plane \(\alpha\). In our case, since \(O\in\alpha\) we may take \((x_{\alpha},y_{\alpha},z_{\alpha})=(0,0,0)\).
We now have \[\mbox{dist }(B,\alpha)=\frac{\left|-5\cdot (-1)-15\cdot 2-4\cdot (-3)\right|}{\sqrt{5^2+15^2+4^2}}=\frac{13}{\sqrt{266}}.\]