Porisma

Work with Cartesian vectors

Connect coordinates, basis vectors, translation and length in two and three dimensions.

Work with Cartesian vectors

For a displacement, subtract the starting coordinates from the ending coordinates. For a length, square the components, add and take the square root. Divide by that length only when a unit direction is needed and the vector is nonzero.

In two dimensions, i and j are unit vectors along the positive x- and y-axes. They are perpendicular:

i=(10),j=(01).\mathbf i=\begin{pmatrix}1\\0\end{pmatrix},\qquad\mathbf j=\begin{pmatrix}0\\1\end{pmatrix}.

For the source point A(4,3), the triangle law gives

OA→=4i+3j=(43),∣OA→∣=42+32=5.\overrightarrow{OA}=4\mathbf i+3\mathbf j=\begin{pmatrix}4\\3\end{pmatrix},\qquad\lvert\overrightarrow{OA}\rvert=\sqrt{4^2+3^2}=5.

Its figure also marks B(2,4), both position vectors and the displacement A→B. In general, OP→=(xy)\overrightarrow{OP}=\begin{pmatrix}x\\y\end{pmatrix} and ∣OP→∣=x2+y2\lvert\overrightarrow{OP}\rvert=\sqrt{x^2+y^2}. Following the path through O gives the complete displacement chain:

AB→=AO→+OB→=−OA→+OB→=OB→−OA→.\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OB}-\overrightarrow{OA}.

The source translation example starts at P(1,4) with v=(3−1)\mathbf v=\begin{pmatrix}3\\-1\end{pmatrix}. Then

OP→+v=(14)+(3−1)=(43)=OQ→.\overrightarrow{OP}+\mathbf v=\begin{pmatrix}1\\4\end{pmatrix}+\begin{pmatrix}3\\-1\end{pmatrix}=\begin{pmatrix}4\\3\end{pmatrix}=\overrightarrow{OQ}.

So Q is (4,3): three units in the positive x-direction and one in the negative y-direction. Addition or subtraction of a fixed displacement translates the point.

Extension: coefficient columns

A coordinate column need not describe a spatial point. In the source extension, a real polynomial a0+a1x+⋯+anxna_0+a_1x+\cdots+a_nx^n can be represented by the (n+1)×1(n+1)\times1 coefficient column (a0a1⋮an)\begin{pmatrix}a_0\\a_1\\\vdots\\a_n\end{pmatrix}, using the fixed ordered basis 1,x,…,xn1,x,\ldots,x^n. This represents polynomials of degree at most n, including zero leading coefficients.

Extend to three dimensions

Use three mutually perpendicular unit directions i,j,k along positive x,y,z. Then

OP→=xi+yj+zk=(xyz),∣OP→∣=x2+y2+z2.\overrightarrow{OP}=x\mathbf i+y\mathbf j+z\mathbf k=\begin{pmatrix}x\\y\\z\end{pmatrix},\qquad\lvert\overrightarrow{OP}\rvert=\sqrt{x^2+y^2+z^2}.

The source example A(4,3,5) gives OA→=(4,3,5)\overrightarrow{OA}=(4,3,5) and ∣OA→∣=42+32+52=52\lvert\overrightarrow{OA}\rvert=\sqrt{4^2+3^2+5^2}=5\sqrt2.

Derive the three-dimensional length

To prove the formulas, the source marks A(x,0,0), B(0,y,0), C(0,0,z), R(x,y,0) and P(x,y,z). In the xy-plane, OR→=xi+yj\overrightarrow{OR}=x\mathbf i+y\mathbf j. Since RP is parallel to the z-axis, RP→=zk\overrightarrow{RP}=z\mathbf k, so

OP→=OR→+RP→=(xi+yj)+zk=xi+yj+zk.\overrightarrow{OP}=\overrightarrow{OR}+\overrightarrow{RP}=(x\mathbf i+y\mathbf j)+z\mathbf k=x\mathbf i+y\mathbf j+z\mathbf k.

Two applications of Pythagoras give

∣OR→∣=x2+y2,∣OP→∣=∣OR→∣2+∣RP→∣2=(x2+y2)2+z2=x2+y2+z2.\lvert\overrightarrow{OR}\rvert=\sqrt{x^2+y^2},\qquad\lvert\overrightarrow{OP}\rvert=\sqrt{\lvert\overrightarrow{OR}\rvert^2+\lvert\overrightarrow{RP}\rvert^2}=\sqrt{(\sqrt{x^2+y^2})^2+z^2}=\sqrt{x^2+y^2+z^2}.

The orthonormal basis columns are i=(100)\mathbf i=\begin{pmatrix}1\\0\\0\end{pmatrix}, j=(010)\mathbf j=\begin{pmatrix}0\\1\\0\end{pmatrix}, k=(001)\mathbf k=\begin{pmatrix}0\\0\\1\end{pmatrix}. For a right-handed system, curl the right-hand fingers from +x towards +y; the thumb points along +z. Viewed from +z towards O, that quarter-turn is anticlockwise. The projected drawing does not show true on-screen lengths or right angles.

Keep the six coordinate results together

For A(x1,y1,z1)A(x_1,y_1,z_1) and B(x2,y2,z2)B(x_2,y_2,z_2):

1. Position vector.

OA→=x1i+y1j+z1k=(x1y1z1)\overrightarrow{OA}=x_1\mathbf i+y_1\mathbf j+z_1\mathbf k=\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}

2. Magnitude.

∣OA→∣=x12+y12+z12\lvert\overrightarrow{OA}\rvert=\sqrt{x_1^2+y_1^2+z_1^2}

3. Unit direction, when A≠O.

OA→∣OA→∣=1x12+y12+z12(x1y1z1)\dfrac{\overrightarrow{OA}}{\lvert\overrightarrow{OA}\rvert}=\dfrac1{\sqrt{x_1^2+y_1^2+z_1^2}}\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}

4. Scalar multiplication.

kOA→=(kx1ky1kz1)k\overrightarrow{OA}=\begin{pmatrix}kx_1\\ky_1\\kz_1\end{pmatrix}

5. Equality of vectors.

OA→=OB→  ⟺  x1=x2, y1=y2, z1=z2\overrightarrow{OA}=\overrightarrow{OB}\iff x_1=x_2,\ y_1=y_2,\ z_1=z_2

All three components must agree. 6. Addition and subtraction.

OA→±OB→=(x1y1z1)±(x2y2z2)=(x1±x2y1±y2z1±z2)\overrightarrow{OA}\pm\overrightarrow{OB}=\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}\pm\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}=\begin{pmatrix}x_1\pm x_2\\y_1\pm y_2\\z_1\pm z_2\end{pmatrix}

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