Vector geometry

Here, we introduce some concepts and notation for the three-dimensional coordinate system. The $3 \mathrm{D}$ coordinate system is often denoted by $\mathbb{R}^{3}$ . A general point $P$ which lives in a 3D space is defined by its $x-, y$ -, $z$ -coordinates. If the point has zero $z$ -coordinate then we say it sits in $x-y$ plane. The $x-y, x-z, y$ -z planes are known as the coordinate planes. Consider a point $P=(x, y, z)$ and a point $Q=(x, y, 0)$ ; the point $Q$ is referred to as the projection of $P$ in the $x-y$ plane.

Much (but not all!) of what we know to be true in $2 \mathrm{D}$ , i.e. in $\mathbb{R}^{2}$ can be extended in $3 \mathrm{D}$ . For instance, the distance between two points $P_{1}=\left(x_{1}, y_{1}\right)$ and $P_{2}=\left(x_{2}, y_{2}\right)$ in $\mathbb{R}^{2}$ is given by

d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}},

and in $\mathbb{R}^{3}$ , for $P_{1}=\left(x_{1}, y_{1}, z_{1}\right)$ and $P_{2}=\left(x_{2}, y_{2}, z_{2}\right)$ ,

d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} .

Vector equation of a line

What information do we need to determine a line? In 2D, we need (i) a point on the line, say $P=\left(x_{1}, y_{1}\right)$ and (ii) a direction or slope, i.e. $k$ . In 3D, we need (i) a point on the line, say $P=\left(x_{1}, y_{1}, z_{1}\right)$ and (ii) a direction given by a vector parallel to the line, say $\boldsymbol{v}$ . In Fig. $9.9 P_{0}$ is a known point on the line $L$ with coordinates $\left(x_{0}, y_{0}, z_{0}\right), P$ is an arbitrary point also on $L$ with coordinate points $(x, y, z)$ . The vectors $\boldsymbol{r}_{\mathbf{0}}$ and $\boldsymbol{r}$ are the position vectors for points $P_{0}$ and $P$ , respectively. Now, $\boldsymbol{a}=\overrightarrow{P_{0} P}$ and by the parallelogram rule (see Section 9.2),

r=r_{0}+a .

The vector $\boldsymbol{v}$ gives us the direction parallel to the line; since $\boldsymbol{a}$ is parallel to $\boldsymbol{v}$ , we can write $\boldsymbol{v}$ as a scalar multiple of $\boldsymbol{a}$ ,

\boldsymbol{a}=\lambda \boldsymbol{v}, \quad \lambda \in \mathbb{R} .

It follows that the line which is parallel to the vector $\boldsymbol{v}$ and passes through the point $P$ with position vector $\boldsymbol{r}$ is given by,

\boldsymbol{r}=\boldsymbol{r}_{0}+\lambda v

We note that $\boldsymbol{r}$ traces out the line, $\boldsymbol{r}_{\mathbf{0}}$ puts us onto the line and $\lambda \boldsymbol{v}$ moves us along the line (if $\lambda>0$ , we move in the direction of $\boldsymbol{v}$ , if $\lambda<0$ , we move in the opposite direction of $\boldsymbol{v})$ .

To write the equation of a line in Cartesian coordinates starting from Eq. (9.25), we write out the vectors in component form

\begin{array}{r} \boldsymbol{r}_{\mathbf{0}}=x_{0} \mathbf{i}+y_{0} \mathbf{j}+z_{0} \mathbf{k}, \\ \boldsymbol{v}=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}, \\ \boldsymbol{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \end{array}

Using Eq. (9.25) and Eqs. (9.26), we write $\boldsymbol{r}-\boldsymbol{r}_{\mathbf{0}}=\lambda \boldsymbol{v}$ as

\left(x-x_{0}\right) \mathbf{i}+\left(y-y_{0}\right) \mathbf{j}+\left(z-z_{0}\right) \mathbf{k}=\lambda(a \mathbf{i}+b \mathbf{j}+c \mathbf{k}) .

Comparing coefficients on each side gives us the following equation

\lambda=\frac{x-x_{0}}{a}=\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c} ;

these are known as the symmetric equations of the line. From Eq. (9.28) we have the following parametric form of the equation of the line,

x=x_{0}+\lambda a, \quad y=y_{0}+\lambda b, \quad z=z_{0}+\lambda c .

Figure 9.9: The vector equation of a line in 3D coordinate system.

Note that if one of $a, b$ , or $c$ is zero in the symmetric equations (9.28), we can still write them down. For instance, suppose $a=0$ ; then, we have

\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c}, \quad x=x_{0}

Example 9.4 Determine the vector equation of the straight line passing through the point $P_{1}$ with coordinates $(3,-1,5)$ and $P_{2}$ with coordinates $(-1,-4,2)$ .

Solution First we obtain the position vector of $P_{1}$ , as $\overrightarrow{O P_{1}}=(3,-1,5)$ . Next, we need the direction given by $\overrightarrow{P_{1} P_{2}}$ as $(-4,-3,-3)$ . The vector equation of a line is then given by

\boldsymbol{r}=3 \mathbf{i}-\mathbf{j}+5 \mathbf{k}-\lambda(4 \mathbf{i}+3 \mathbf{j}+3 \mathbf{k})

where $\lambda \in \mathbb{R}$ .

Vector equation of a plane

Here we derive a general equation for planes. We start with a point $P_{0}$ which is known to be on the plane, as shown in Fig. 9.10 . The point $P$ is an arbitrary point on the plane and $\boldsymbol{r}$ and $\boldsymbol{r}_{\mathbf{0}}$ are the position vectors of $P$ and $P_{0}$ , as before. We also show a vector, $\boldsymbol{n}$ which is perpendicular to the plane; this is known as the normal vector. Notice that the vector $\boldsymbol{r}-\boldsymbol{r}_{\mathbf{0}}$ lies completely in the plane. Since $\boldsymbol{n}$ is orthogonal to the plane, it is also orthogonal to any vector that lies in the plane. Using the dot product, we have $\boldsymbol{n} \cdot\left(\boldsymbol{r}-\boldsymbol{r}_{\mathbf{0}}\right)=0$ which gives

n \cdot r=n \cdot r_{0}

Figure 9.10: The vectors $\boldsymbol{r}$ and $\boldsymbol{r}_{\mathbf{0}}$ are position vectors, the points $P_{0}$ and $P$ are on the plane and $\boldsymbol{r}-\boldsymbol{r}_{\mathbf{0}}$ lies completely in the plane. Also, $\boldsymbol{n}$ is the normal vector which is perpendicular to the plane.

as the vector equation of the plane.

Starting from $\boldsymbol{n} \cdot\left(\boldsymbol{r}-\boldsymbol{r}_{\mathbf{0}}\right)=0$ with $\boldsymbol{n}=(a, b, c), \boldsymbol{r}=(x, y, z)$ , and $\boldsymbol{r}_{\mathbf{0}}=\left(x_{0}, y_{0}, z_{0}\right)$ we have

(a, b, c) \cdot\left(x-x_{0}, y-y_{0}, z-z_{0}\right)=0

which yields

a\left(x-x_{0}\right)+b\left(y-y_{0}\right)+c\left(z-z_{0}\right)=0,

known as the scalar equation of plane. This is often written as

a x+b y+c z=d,

where $d=a x_{0}+b y_{0}+c z_{0}$ .

Alternatively, we may also specify a plane by a point on the plane, $P_{0}$ with position vector $\boldsymbol{r}_{\mathbf{0}}$ and two direction vectors, say $\boldsymbol{u}$ and $\boldsymbol{v}$ which are not parallel or anti-parallel. In this way, the definition is similar to that describing the vector equation of the line. This is written in parametric form as

\boldsymbol{r}=\boldsymbol{r}_{0}+\lambda \boldsymbol{u}+\mu \boldsymbol{v}

where $\lambda, \mu \in \mathbb{R}$ . Example 9.5 Give the equation of the plane containing the points $(0,0,2)$ , $(0,2,3)$ , and $(1,1,1)$ in vector and Cartesian form.

Solution We can start with the vector form given by Eq. (9.33). We define the position vector $\boldsymbol{r}_{\mathbf{0}}$ is using the point $(0,0,2)$ such that $\boldsymbol{r}_{\mathbf{0}}=(0,0,2)$ . Next, since $(0,0,2),(0,2,3)$ , and $(1,1,1)$ all lie on the plane, then any vector between them is also in the plane. We define

\boldsymbol{u}=(0-0,2-0,3-2)=(0,2,1), \quad \boldsymbol{v}=(1-0,1-0,1-2)=(1,1,-1)

The equation of the plane is then defined by

\boldsymbol{r}=(0,0,2)+\lambda(0,2,1)+\mu(1,1,-1)

This can also be expressed as

x=\mu, \quad y=2 \lambda+\mu \quad z=2+\lambda-\mu .

Now, for the scalar form, since the direction vectors $\boldsymbol{u}, \boldsymbol{v}$ lie in the plane then $\boldsymbol{n}$ , which is normal to the plane, is defined by taking the cross product of the direction vectors. This is given by

\boldsymbol{n}=\boldsymbol{u} \times \boldsymbol{v}=-3 \mathbf{i}+\mathbf{j}-2 \mathbf{k}

Finally, taking the dot product as in Eq. (9.31) with $a=-3, b=1, c=-2$ and $\left(x_{0}, y_{0}, z_{0}\right)=(0,0,2)$ , we obtain the Cartesian form of the equation of the plane as,

(-3,1,-2) \cdot(x, y, z-2)=-3 x+y-2 z-4=0,

yielding

3 x-y+2 z=4 .

Cross/vector product Coplanar points