Component form of r

If $\bar r$ is the position vector of a point $P$ with respect to the origin, then $\bar r=x\hat\imath +y\hat\jmath+z\hat k$, and $x,y,z$ are called the components of $\bar r$ along the $X,Y,Z$ axes; every

vector in space is a unique combination of $\hat\imath,\hat\jmath,\hat k$ this way (some texts write this

using angle brackets, $\bar r=\langle x,y,z\rangle$, for the same point $(x,y,z)$).

Two vectors $\bar a=a_1\hat\imath+a_2\hat\jmath+a_3\hat k$ and $\bar b=b_1\hat\imath+b_2\hat\jmath+b_3\hat k$

are equal exactly when their like components agree, $a_1=b_1,\ a_2=b_2,\ a_3=b_3$; their sum is found by

adding like components, $\bar a+\bar b=(a_1+b_1)\hat\imath+(a_2+b_2)\hat\jmath+(a_3+b_3)\hat k$; and for a

scalar $k$, $k\bar a=ka_1\hat\imath+ka_2\hat\jmath+ka_3\hat k$. If $\bar b=k\bar a$ (collinear vectors), then

$b_1/a_1=b_2/a_2=b_3/a_3=k$. The unit vector along $\bar a$ is

$\hat a=\bar a/|\bar a|=\dfrac{a_1\hat\imath+a_2\hat\jmath+a_3\hat k}{\sqrt{a_1^2+a_2^2+a_3^2}}$.

Three further standing results (used repeatedly in the exercises) are stated here: (v) three distinct points

with position vectors $\bar a,\bar b,\bar c$ are collinear iff there exist non-zero scalars $x,y,z$ with

$x\bar a+y\bar b+z\bar c=\bar 0$ and $x+y+z=0$; (vi) four points (no three collinear) with position vectors

$\bar a,\bar b,\bar c,\bar d$ are coplanar iff there exist scalars $x,y,z,w$, not all zero, with …