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3.4 Vectors (HL)

Tags
Vectors
Line
Plane
Intersection
Distance
Angle
Dot product
Cross product
Cartesian
Parametric
Terminology
Definition
Scalar
Quantities which have only magnitudes.
Vector a→\overrightarrow{a}
Quantities which have both direction and magnitude.
They must be written with the arrow on top; the printed text represents them in bold.
Figure 3.4.1 Vector a→\overrightarrow a , vector AB→\overrightarrow {AB} and BA→\overrightarrow {BA}
The magnitude of vector is the length of that vector:  ∣a→∣|\overrightarrow{a}|
The negative of a vector a→\overrightarrow{a} is in the opposite direction of a→\overrightarrow a, but has the same length.
Two vectors are equal if they have the same direction and magnitude.
Zero vector 0→\overrightarrow 0 is a vector of length 00.
Vector addition
a→+b→\overrightarrow a+\overrightarrow b
Figure 3.4.2 Addition of two vectors
Geometrically, join the endpoint of a→\overrightarrow a with the starting point of b→\overrightarrow b.
AB→+BC→=AC→\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}
Vector subtraction
a→−b→\overrightarrow a-\overrightarrow b
Figure 3.4.3 Difference of two vectors
Geometrically, simply add −b→-\overrightarrow b to a→\overrightarrow a like above.
AB→=OB→−OA→\overrightarrow{AB}=\overrightarrow {OB}-\overrightarrow{OA}
Unit vector
Unit vectors have a magnitude of 1.
You can obtain the unit vector in the a→\overrightarrow a direction by: 1∣a→∣a→\frac{1}{|\overrightarrow a|}\overrightarrow a
Basis vector
The unit vector which spans the entire vector space.
In 2 dimension, the basis vectors are: i→=(10)\overrightarrow i=\binom 10, j→=(01)\overrightarrow j=\binom01
In 3 dimensions, the basis vectors extend to: i→=(100)\overrightarrow i=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} , j→=(010)\overrightarrow j=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} , k→=(001)\overrightarrow k=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
Column vector
They represent the relative position to the origin.
If we have A(1,3,−2)A(1, 3, -2), then OA→=(13−2)\overrightarrow {OA}=\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}
Vector Algebra:
1.
(ab)±(cd)=(a±cb±d)\binom ab\pm\binom cd= \binom {a \pm c}{b \pm d}
2.
k(ab)=(kakb)k\binom ab=\binom {ka}{kb}
3.
∣(ab)∣=a2+b2|\binom ab|=\sqrt {a^2+b^2}
These can be extended to nnth dimension.
a→//b→\overrightarrow a // \overrightarrow b iff a→=kb→\overrightarrow a=k\overrightarrow b.
Inner product
Dot product
a→⋅b→=∣a→∣∣b→∣cos⁡θ\overrightarrow a · \overrightarrow b = |\overrightarrow a| |\overrightarrow b|\cos\theta
(ab)⋅(cd)=ac+bd\binom ab \cdot \binom cd=ac+bd
Note that the result of this operation is a scalar.
You can rearrange this to obtain the angle between the two vectors.
a→⊥b→\overrightarrow a ⟂\overrightarrow b iff a→⋅b→=0\overrightarrow a ·\overrightarrow b = 0
Outer product
Cross product
(HL)
∣a→×b→∣=∣a→∣∣b→∣sin⁡θ|\overrightarrow a \times \overrightarrow b| =|\overrightarrow a | |\overrightarrow b|\sin\theta
Note that the result of this operation is a vector.
The geometric interpretation of this operation is the vector perpendicular to both a→\overrightarrow a, b→\overrightarrow b.
Figure 3.4.4 Difference between a→×b→\overrightarrow a \times \overrightarrow b and b→×a→\overrightarrow b \times\overrightarrow a
Therefore, the direction matters, and a→×b→=−b→×a→\overrightarrow a \times \overrightarrow b = -\overrightarrow b \times\overrightarrow a. (not commutative)
Components
(HL)
It follows that:
1.
Component of a→\overrightarrow a in the direction of b→\overrightarrow b is: a→⋅b→∣b→∣\frac {\overrightarrow a ·\overrightarrow b}{|\overrightarrow b|}
2.
Component of a→\overrightarrow a perpendicular to b→\overrightarrow b is ∣a→×b→∣∣b→∣\frac {|\overrightarrow a \times\overrightarrow b|}{|\overrightarrow b|}

Vector Applications

Types
Line Equation
Steps
We can now define the linear equation in multiple ways:
We call d→\overrightarrow d as the direction vector.
The vector equation of a line has a form: r→=a→+λb→ \overrightarrow r =\overrightarrow a +\lambda\overrightarrow b, (xy)=(ab)+λ(d1d2)\binom xy=\binom ab+\lambda\binom {d_1}{d_2}
The parametric equation of a line has a form: x=a+λd1x = a + \lambda d_1, y=b+λd2y = b + \lambda d_2.
The cartesian equation of a line has a form: ax+by+c=0ax + by + c =0
Relationship:
1.
l1//l2l_1 // l_2 iff d1→//d2→\overrightarrow {d_1}//\overrightarrow {d_2}
2.
l1⊥l2l_1 ⟂ l_2 iff d1→⊥d2→\overrightarrow {d_1} ⟂\overrightarrow{ d_2}
3.
If l1l_1 and l2l_2 have a unique intersection, solve the simultaneous equation with their parametric equation.
4.
cos⁡θ=∣d1⋅d2∣∣d1∣∣d2∣\cos\theta =\frac {|d_1 · d_2|}{|d_1||d_2|}
5.
l1l_1 and l2l_2 are skew if they are not parallel, and do not intersect.(3D)
Note that this can be extended to 3D by adding the zz coordinate.
Area
A=12∣a→∣∣b→∣sin⁡θ=12∣a→×b→∣A = \frac12|\overrightarrow a||\overrightarrow b| \sin\theta= \frac12|\overrightarrow a \times \overrightarrow b|
Kinematics(const. velocity)
The direction vector is the velocity vector.
It must have the magnitude equal to the given velocity.
Shortest distance
This is finding the shortest distance given a point P(a,b)P(a, b) and a line ll.
1.
Find the direction vector d→\overrightarrow d of ll
2.
Parametrize the foot of perpendicular, NN on the line ll.
3.
Find PN→\overrightarrow {PN}.
4.
Using PN→⊥d→\overrightarrow {PN} ⟂ \overrightarrow d , find NN.
5.
Calculate PN→\overrightarrow {PN} using the distance formula.
To check this quickly, you could also use a different formula.
Given a point P(x0,y0)P(x_0, y_0) and l:ax+by+c=0l : ax + by + c = 0,
distance=∣ax0+by0+c∣a2+b2distance =\frac{ |ax_0+ by_0 + c|}{\sqrt {a^2 + b^2}}
Plane equation
We can also define planes in 3D.
Plane equation can be determined by two coplanar vectors, or one normal vector.
The vector equation of a plane has the form: r→=a→+λv1→+µv2→\overrightarrow r= \overrightarrow a+\lambda \overrightarrow {v_1}+µ\overrightarrow {v_2},
(xyz)=(abc)+λ(a1b1c1)+µ(a2b2c2)\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} a \\ b \\ c \end{pmatrix}+ \lambda\begin{pmatrix} a_1 \\ b_1 \\ c_1 \end{pmatrix}+µ\begin{pmatrix} a_2 \\ b_2\\ c_2\end{pmatrix}
The cartesian equation has the form:
n→⋅r→=0\overrightarrow n · \overrightarrow r=0, i.e. if n→=(abc)\overrightarrow n=\begin{pmatrix} a \\ b \\ c \end{pmatrix} , and r→=(x0y0z0)\overrightarrow r=\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} , we have:
a(x−x0)+b(y−y0)+c(z−z0)=0a(x-x_0)+b(y-y_0)+c(z-z_0)=0
Note that you can find n→\overrightarrow n as a cross product of any two coplanar vectors.
Relationship:
1.
Π1//Π2\Pi_1 // \Pi_2 if n1→//n2→\overrightarrow {n_1}//\overrightarrow {n_2}
2.
Π1⊥Π2\Pi_1 ⟂\Pi_ 2 if n1→⊥n2→\overrightarrow {n_1} ⟂ \overrightarrow {n_2}
3.
We can construct a system of equations using the parametric equations of Π1\Pi_1, Π2\Pi_2
a.
If the system is consistent, it intersects on a unique point.
b.
Otherwise, it intersects on a line.
4.
Angle between two planes: cos⁡θ=∣d1⋅d2∣∣d1∣∣d2∣\cos \theta =\frac{|d_1 · d_2|}{|d_1||d_2|}
5.
Angle between a plane and a line: sin⁡θ=∣n⋅d∣∣n∣∣d∣\sin\theta=\frac{|n · d|}{|n||d|}
Shortest distance
This is finding the shortest distance given a point A(a,b,c)A(a, b, c) and a plane Π\Pi.
Figure 3.4.5 Shortest distance in a plane (3D)
1.
Find the normal vector of the plane Π\Pi
2.
Parameterize the line AN→\overrightarrow {AN}
3.
As the line intersects the plane at point NN, substitute the parametric form to obtain the chosen parameter, λ\lambda
4.
Calculate ∣AN→∣|\overrightarrow {AN}|
To check this quickly, you could also use a different formula.
Given a point P(x0,y0,z0)P(x_0, y_0, z_0) and Π\Pi: ax+by+cz+d=0 ax + by + cz +d= 0,
distance=∣ax0+by0+cz0+d∣a2+b2+c2distance = \frac{|ax_0+ by_0 + cz_0+d|}{\sqrt{a^2 + b^2+c^2}}
3.4 Vectors
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3.4 Vectors (HL)