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Wize University Calculus 3 Textbook > Vectors & Geometry of Space

Vectors in 2-dimensional (2D) and 3-dimensional (3D) Systems

Properties of Vectors, Length of Vectors, and Unit VectorsNext Section
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Vectors in 2D & 3D Systems


Vectors in 2-D and 3-D coordinate systems show a quantity that has a direction and magnitude, vs a scalar quantity which has just magnitude.

Vectors do not tell us information about where the quantity was applied in space.

2-D space:

  • The vector v⃗\vec{v} is shown as:
v⃗=<a1, a2>=a1i⃗+a2j⃗\vec{v}=\rowvec{a_1}{a_2} = a_1\vec{i}+a_2\vec{j}
where a1,a2 a_1,a_2~ are the components of the vector and i⃗\vec{i} and j⃗\vec{j} , which are the basic unit vectors in xx and yy directions, respectively.

  • These unit vectors are:
i⃗=<1, 0, 0>j⃗=<0, 1, 0>\begin{array}{l} \vec{i} = \rowvecth{1}{0}{0}\\[5pt] \vec{j} = \rowvecth{0}{1}{0}\\[5pt] \end{array}

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3-D space:

  • The vector v⃗\vec{v} is shown as:
v⃗=<a1, a2, a3>=a1i⃗+a2j⃗+a3k⃗\vec{v}=\rowvecth{a_1}{a_2}{a_3} = a_1\vec{i}+a_2\vec{j}+a_3\vec k
where a1,a2,a3 {a_1}, {a_2}, {a_3}~are the components of the vector and i⃗\vec{i}, j⃗\vec{j}, and k⃗ \vec{k}~ are the basic unit vectors in xx, yy, and z z~ directions, respectively.
  • These unit vectors are:
i⃗=<1, 0, 0>j⃗=<0, 1, 0>k⃗=<0, 0, 1>\begin{array}{l} \vec{i} = \rowvecth{1}{0}{0}\\[5pt] \vec{j} = \rowvecth{0}{1}{0}\\[5pt] \vec k = \rowvecth{0}{0}{1}\\ \end{array}

Need to Know:
  • A unit vector is a vector of magnitude = 1
  • A position vector starts at the origin and ends at a point P = (x, y, z)
  • Given two points, A=(a1, a2, a3)A=(a_1,\ a_2,\ a_3) and B=(b1, b2, b3)B=\left(b_1,\ b_2,\ b_3\right), the vector AB⃗\vec{AB} is:
AB⃗=v⃗=<b1−a1, b2−a2, b3−a3>\boxed{\vec{AB} = \vec{v}= \rowvecth{b_1-a_1}{b_2-a_2}{b_3-a_3}}

Watch Out!
Direction is important!

The vector BA⃗\vec{BA} is:

BA⃗=v⃗=<a1−b1, a2−b2, a3−b3>\vec{BA} = \vec{v}= \rowvecth{a_1-b_1}{a_2-b_2}{a_3-b_3}

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Example


Let X=(2, −7, 0)X=\left(2,\ -7,\ 0\right) and Y=(1, −3, −5)Y=\left(1,\ -3,\ -5\right).


a) Find XY⃗\vec{XY}

To find XY⃗\vec{XY}, we must take the coordinates points for Y and subtract the coordinates of X from them. Thus,
XY⃗= <1−2, −3−(−7), −5−0>XY⃗= <−1, 4, −5>\begin{array}{rcl} \vec{XY}&=&~<1-2,\ -3-\left(-7\right),\ -5-0>\\\\ \vec{XY}&=&~<-1,\ 4,\ -5> \end{array}



b) Find YX⃗\vec{YX}

To find YX⃗\vec{YX}, we will switch the order of subtraction for our coordinate points.
YX⃗= <2−1, −7−(−3), 0−(−5)>YX⃗= <1, −4, 5>\begin{array}{rcl} \vec{YX}&=&~<2-1,\ -7-\left(-3\right),\ 0-\left(-5\right)>\\\\ \vec{YX}&=&~<1,\ -4,\ 5> \end{array}

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Example


Suppose A=(a1, a2, a3)A=(a_1,~a_2,~a_3) and B=(b1, b2, b3)B=(b_1,~b_2,~b_3).

Let v⃗= <b1−a1, b2−a2, b3−a3>, \vec{v}=~<b_1-a_1,~b_2-a_2,~b_3-a_3>,~where a1, a2, a3 a_1,~a_2,~a_3~are the components of AA and b1, b2, b3 b_1,~b_2,~b_3~are the components of BB.

Which of the following is BA⃗\vec{BA}?

a)  v⃗~\vec{v}

b)  −v⃗~\vec{-v}

Since v⃗= <b1−a1, b2−a2, b3−a3> \vec{v}=~<b_1-a_1,\ b_2-a_2,\ b_3-a_3>~ represents all the components of BB minus all the components of AA, then v⃗=AB⃗\vec{v}=\vec{AB}.

Therefore, −v⃗=BA⃗\vec{-v}=\vec{BA}.

So, b).

Practice

Suppose A=(a1,a2,a3)A=(a_1, a_2, a_3) and B=(b1,b2,b3)B=(b_1, b_2, b_3).

If AB⃗= <3,−9,1>, \vec{AB}=~<3, -9, 1>,~what is BA⃗?\vec{BA}?

Practice

True or False:

  • If α⃗=(α1, α2, α3), ϕ⃗=(ϕ1, ϕ2, ϕ3)\vec{\alpha}=(\alpha_1,~\alpha_2,~\alpha_3),~\vec{\phi}=(\phi_1,~\phi_2,~\phi_3) and
  • v⃗= <α1−ϕ1, α2−ϕ2, α3−ϕ3>, \vec{v}=~<\alpha_1-\phi_1,~\alpha_2-\phi_2,~\alpha_3-\phi_3>,~
  • then −v⃗=ϕα⃗\vec{-v}=\vec{\phi\alpha}