Vector

Let $\QTR{bf}{R}^3$ denote the set of all triples $(a_1,a_2,a_3)$ with MATH Elements in $\QTR{bf}{R}^3$ are called vectors. The sum of vectors
MATH
is
MATH
The scalar multiplication of MATH with the vector $a=(a_1,a_2,a_3)$ is
MATH
The zero vector is denoted $0=(0,0,0).$ For MATH $-a$ is defined by $-a=(-1)a.$ These rules have properties

Geometric Interpretation

translation: $x\rightarrow x+a$

magnification: MATH $\alpha >1$

contraction: MATH

parallelogram construction

Dot Product

The dot product of two vectors $a=(a_1,a_2,a_3)$ and $b=(b_1,b_2,c_3)$ is
MATH

Proposition

Cauchy-Schwarz Inequality


MATH

The equality holds if and only if there exist MATH with MATH

Proof. Case 1. $a=0.$ Then MATH so the equality holds.

Case 2. $a\ne 0.$ Then $a\cdot a>0.$ Consider the vector
MATH
Then
MATH

MATH

MATH
Therefore MATH

If MATH and $a\ne 0,$ then MATH so $c=0;$ i.e. MATH If MATH and $b\ne 0,$ then MATH

The norm of the vector $a=(a_1,a_2,a_3)$ is
MATH
It follows from Cauchy-Schwarz inequality that
MATH
if $a\ne 0,b\ne 0.$ The unique angle MATH such that
MATH
is called the angle between the vectors $a$ and $b.$ This is consistent with the Law of Cosine:
MATH

MATH

MATH

MATH

Properties of the Norm

Proof.
MATH

MATH

MATH

MATH

MATH

Two vectors $a,b$ are perpendicular $(a\perp b)$ if the angle between them equals $\pi /2:$
MATH

Projections and Components

Theorem. Let $a\ne 0.$ Each $b\in \QTR{bf}{R}^3$ can be uniquely expressed as
MATH
where $b_{\Vert }$ is parallel to $a$ (i.e., there exists MATH with MATH and $b_{\perp }$ is perpendicular to $a$ (i.e., MATH

Proof. Set MATH and MATH Then
MATH
If $b=b_1+b_2$ with MATH then
MATH
so MATH; this shows MATH and MATH

Definition MATH is called the projection of the vector $b$ on the vector $a.$

Direction Cosine

Notation: MATH

If $a\ne 0,$ then the vector MATH satisfies MATH i.e., $u$ is a unit vector. The components of $u $ in the directions $i,j,k$
MATH

MATH

MATH
are called the direction cosines of $a.$

Cross Product

Let MATH Their cross product (also called vector product) is given by
MATH

Properties of Cross Product

Cramer's Rule

Consider the simultaneous set of equations
MATH

MATH

MATH
From the basic properties of the determinant we have
MATH
Hence if $x$ is to satisfy equations (1) then
MATH
Write
MATH
If $\Delta \ne 0,$ then
MATH
Similarly, if $\Delta \ne 0,$ then
MATH

On the other hand, the following identity always holds:
MATH
This four-by-four determinant can be expanded along the first row:
MATH

MATH
Hence if $\Delta \ne 0,$ then
MATH
is indeed the (unique) solution to the system (1).

The method can be extended to $n$ linear equations in $n$ unknowns.

Straight Line in $\QTR{bf}{R^3}$

A straight line in $\QTR{bf}{R^3}$ can be given by the parametric equations:
MATH
A point $(x,y,z)$ belong to this line if and only if
MATH
for some $t\in \QTR{bf}{R.}$ This can therefore be written in vector equations as:
MATH
After eliminating the parameter $t$ from the parametric equation, we obtain the symmetric equation
MATH
if $abc\ne 0.$

Example Find the equations of the straight line passing through $p=(-1,4,3)$ and $q=(2,-1,5)$.

First we determine the direction: $q-p=(3,-5,2);$ from this the vector equation is obtained:
MATH

MATH
Equating the components we obtain the parametric equation:
MATH
and the symmetric equations:
MATH

Example Find the distance between the skew lines $L_1$ given by
MATH
and $L_2$ given by
MATH

We note that $L_1$ has direction $u=(1,-1,-2)$ and $L_2$ has direction $v=(-2,2,-2).$ The vector
MATH
is perpendicular to both $u$ and $v.$ Choose any point $p\in L_1$ and any point $q\in L_2.$ (There are infinitely many ways to do so.) The distance between $L_1$ and $L_2$ is the norm of the projection of $p-q$ on $a.$ Take $p=(-1,2,-1)$ and $q=(1,2,2).$ Then $p-q=(-2,0,-3)$ and so the required distance is
MATH

Plane in $\QTR{bf}{R^3}$

Let $(x_0,y_0,z_0)$ be any point of the plane. The plane consists of all the points MATH satisfying the vector equation
MATH

MATH
for a nonzero vector $n=(a,b,c).$ This vector $n$ is called the normal of the plane. Thus the equation of the plane takes the form
MATH
where
MATH

Example Find the distance from $p=(-1,4,5)$ to the plane $3x-2y+4z=12.$

The vector $n=(3,-2,4)$ is normal to the plane. Choose a point on the plane: $(0,0,3).$ the distance from $p$ to the plane is the norm of the projection of the vector $(-1,4,5)-(0,0,3)$ on $n=$
MATH

Example Show that the line
MATH
can be expressed as the intersection of two planes, each of which is parallel to a coordinate axis.

Consider the plane $P_1$ with equation obtained from the first equality
MATH
$P_1$ can be expressed by the equation
MATH
which has $(4,-2,0)$ as normal. $P_1$ is therefore parallel to the $z$-axis since $(-4,-2,0)$ is perpendicular to $k.$ Similarly, the plane satisfying the equation
MATH
which is the same as the equation
MATH
has $(0,1,-4)$ as normal, must be parallel to the $x$-axis since $(0,1,-4)$ is perpendicular to $i.$ The given line is the intersection of $P_1$ and $P_2.$

Intersecting Planes

The angle between two intersecting planes is the same as the angle between their normals $n_1,n_2$. Depending on the choices of the normals, there are two such angles, each the supplement of the other. We choose the smaller angle, the one with the nonnegative cosine:
MATH

Example Find the cosine of the angle between the planes with equations
MATH
and
MATH

Solution: MATH; so
MATH

Example The planes
MATH
and
MATH
intersect to form a line. Find the vector equation of this line.

Solution: The direction vector of the intersecting line must be perpendicular to both normal vectors $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2).$ Hence it is parallel to MATH Let $(x_0,y_0,z_0)$ be any point on the intersecting line. The required vector equation takes the form
MATH

Plane Determined by Three Noncollinear Points

Suppose that the points
MATH
do not lie on a straight line. A point $x=(x_1,x_2,x_3)$ lies on the plane determined by these three points if and only if the volume of the parallelepiped with $a,b,c,x$ as vertices is zero. In view of the vector triple product identity, the condition is given by
MATH

Example Find the equation for the plane passing through
MATH

Solution: Expanding the determinant
MATH
we have
MATH

Distance from a Point to a Plane

Suppose that $p=(x_1,y_1,z_1)$ does not lie on the plane
MATH
Choose any point $q=(x_0,y_0,z_0)$ on the plane. The distance from $p$ to the plane is the norm of the projection of $p-q$ on the normal $(a,b,c)$ of the plane:
MATH
Since $(x_0,y_0,z_0)$ lies on the plane, it follows that
MATH
Therefore
MATH
consequently the required distance is given by
MATH

Exercise

  1. Find the equation for the plane

    1. through $(1,2,-1)$ perpendicular to $i+j.$

    2. through $(1,2,-1)$ perpendicular to $i+2j-k.$

    3. through $(1,0,1)$ parallel to $x+2y+z=0.$

    4. through MATH parallel to $x+y+z=1.$

  2. Explain why a plane cannot

    1. contain $(1,2,3)$ and $(2,3,4)$ and be perpendicular to $i+j$

    2. contain MATH and $(1,1,1)$

    3. go through the origin and have the equation $ax+by+cy=1$

  3. Find the projection $P$ of $B$ along $A$ and also find $\Vert P\Vert $

    1. MATH

    2. MATH

    3. $B=$unit vector at $60^{\circ }$ angle with $A$

    4. $B=$vector of length $2$ at $60^{\circ }$ angle with $A$

    5. $B=-A$

    6. $A=i+j,B=i+k$

    7. $A$ is perpendicular to $x-y+z=0,B=i+j$

    8. $A$ is perpendicular to $x-y+z=5,B=i+j+5k$

  4. True or false:

    1. $A\times B$ never equals $A\cdot B.$

    2. If $A\times B=0$ and $A\cdot B=0,$ then either $A=0$ or $B=0.$

    3. If MATH and $A\ne 0,$ then $B=C.$

    4. MATH

    5. MATH

    6. MATH

  5. Which of the following equals to $A\times B?$

    1. $(A+B)\times B$

    2. $(-B)\times (-A)$

    3. MATH

    4. $(A+C)\times (B-C)$

    5. MATH