LINeS What information do we reed to desuike, a straight line in R'? We weed tio thags: = Any print Py (ts)y,, 24) on the ne ~ Ary vector vz Ca,b,c? which is parallel to the line. How dees this work? For each choice af the parameter t we get a vector Lit) which points to a point on the line. We call eaation for Le Cl) = 2,4 tho vextor We can also focus on the components of rit) as follons: ce Quy 2 Here x,y, and & depend on t. They ore fa, ta ' 3 cot Mxy zy These numbers are constants. a rte Se vit dabeyThus we have — 49,27 = Cay, 207 + £ Cabjcy = Casta, yo rth, @ +te7 By comparing components we obtain the parametric, equations of L: xy t to, yey + tb, 2 4 te ExaMece 1 a) Find a vector equation and parametric equations for the line which goss thrugh the point (5,13) and is paralle] to the vector i+ 4p -ak ©) Find ang two other prints on the line. SOLUTION af, = 6543) vt 4y 4-27 has we bove c= cpr ty = CSiay t+ eXya,-a) =Q ste, tus, a-al) fosomekic Equations: = xe Sth, y=ltye, ge 3-ab teR WTR bet thn co = C4,73,57 ee then cud = C4SID Other twe points: — (4,-3,5) and (6,5,1). The wetkor ui called the direction vector (or jut the direction) of the line. Two tines are called parallel i their direction vectors ar parallel , ie. when one is a scalar multiple of the other. If two lines are neither paralel nor intersecting, then they are called skew. There is a third way to describe a line. This i done by climinating the parameter + from the parametric equations, as seen below HO 2st tta , yey, es atee Kets go Yo z- 2 then teary exp) te and thus, X= fe 37 Yo eeThe equations above are called the symmetric equabions of the line. LINE Seq@Menrs. We can also Focus on a finite piece of a straight bine. This is called o li te. The vector equation for a line segment from C, to ©, is given s ii 3 Le Uee)r oe eo, where ost Ss) LE teo then c = (i-o)fo, + 0%, = © Te tei then co = (the, t if, = © Ly Thus 05 the parameter + (we often think of the parameber as time”) changes from 0 to |, the veckor f moves nearly From £, & c. ExAMeLe 2 Find 0 vector equation for the line segment from (a,-1,4) be (y,b,1), Hence weike down the parametric equations fir the same line segment, SoLUTion cee Gaytuy ond o£, = Cust? therefore. cel-t)e + eq te Cai = leeddanuy + eC yg 1? = Capst, 2b-1, yD Facametric eaysations: x = at at) ge Fe-1, rey-3t, telPLANES To descibe 0 lane we also reed a print on the plane and a direction vector. £ > A vector describing 0 gereral ports on “the plane £,—> A vector pointing to the Lown point on the plane. fh —> The dicecion vector, We tall it the normal vector. The veckor ¢ points te a point on the plane precisely when (ce - c,) Lm , that is, when This is called the vector equation of the plane. Two planes are lel if their normal veckors are parallel Te a = (abe Ly = (Le 4 Bo doo te rye, ‘then - a Pe nee ts (ue? (abe? + Cx- x, JoJo Bo 2) = 0 alt- x6) + b(y-y) + cle-2.) = 0 This is called the scalar equation of the plane. Multiplying out the. brackets , we hain the equivalent version fae aig wt at by tte ad where d= 2% + by, +620.EXAMPLE 3 Find the equation of the plant which theragh the points Ply), duu), and "els2.0) eens SoLution. Pes Aa - Gay = Cab) R= Gay - Gy3,27 = Cy +a? Hane eb n= fax ie = 2 k 2 4 4 2 eal) = (Sue -C-y-w)d + (-ay Wk = (lb + aot + hk Neutor exation (c - Gua) + (nau) = 0 Scalar equation: aes) + ao(y-s) + u(e-a) <0