Example 0.1.Vector equation of a line

1 Example vector equation of a a vector equation for the line through(4,6, 3)and parallel tov= 5i 10j+ :With the identificationsx0= 4, y0= 6, z0= 3, a= 5, b= 10 andc= 2 we obtain from< x, y, z >=< x(t), y(t), z(t)>=< x0+at, y0+bt, z0+ct >=< x0, y0, z0>+t < a, b, c >,a vector equation of the line :< x, y, z >=<4,6 ,3>+t <5, 10,2> . Example vector equation of a a vector equation for the line though(2, 1,8)and(5,6, 3).Solution:If we label the points asP0(2, 1,8) andP1(5,6, 3), then a direction vector for the line throughP0andP1isv=# P0P1=# OP1 # OP0=<5 2,6 ( 1), 3 8>=<3,7, 11> .From the vector equation of that liner=r0+t(r1 r0),(herer=# OP ,r0=# OP0,r1=# OP1) a vector equation of the line is< x, y, z >=<2, 1,8>+t <3,7, 11>.

2 This is one of many possible equations of the line . For Example , two alternative equations are< x, y, z >=<5,6, 3>+t <3,7, 11>,< x, y, z >=<5,6, 3>+t < 3, 7,11> . Example Parametric equation of a a parametric equation for the line through(5,2,4) parallel tov= 4i+ 7j :Make identificationsx0= 5, y0= 2, z0= 4, a= 4, b= 7 andc= 9 so that from the parametricequationx(t) =x=x0+at, y(t) =y=y0+bt, z(t) =z=z0+ct,we getx= 5 + 4t, y= 2 + 7t, z= 4 linesL1andL2with direction vectorsv1andv2, respectively, arei) perpendicular ifv1 v2= 0, andii) parallel ifv2=kv1for some non-zero perpendicular whether the linesL1:x= 6 t, y= 20 + 3t, z= 1 + 2t( )L2:x= 5 + 2s y= 9 4s, z= 1 + 7s( )are :Reading off the coefficients of the parameterstands, we see thatv1= i+ 3j+ 2kandv2= 2i 4j+ 7kare the direction vectors forL1andL2.

3 Becausev1 v2= 2 12 + 14 = 0, we conclude that thelines are parallel whether the linesL1:x= 4 2t, y= 1 + 4t, z= 3 + 10t( )L2:x=s y= 6 2s, z=12 5s( )are :Reading off the coefficients of the parameterstands, we see thatv1= 2i+ 4j+ 10kandv2=i 2j 5kare the direction vectors forL1andL2. Becausev1= 2v2, we conclude that the lines are equation of a an equation of the plane that contains the point(4, 1,3)and is perpendicular to the vectorn= 2i+ 8j :It follows immediately from the equation of the plane containingP0(x0, y0, z0) and with normalvectorn=ai+bj+ck, that is,a(x x0) +b(y y0) +c(z z0) = 0,that the identificationx0= 4, y0= 1, z0= 3, a= 2, b= 8, c= 5 yields2(x 4) + 8(y+ 1) 5(z 3) = 0 Example Normal vector a graph of a linear equationax+by+cz+d= 0, witha, b, cNOT all zero, is a plane withnormal vectorn=ai+bj+ instance by reading off the coefficientsx, y, zin the linear equation3x 4y+ 10z 8 = 0,we obtain the normal vectorn= 3i 4j+ 10kOf course a non-zero scalar multiple of a normal vectornis still perpendicular to the non collinear pointsP1, P2, P3also determine a planeS.

4 To obtain an equation of theplane, we need only form two vectors between two pairs of the points. The cross product is a vectornormal to the plane containing these vectors. IfP(x, y, z) represents any point on the plane and3r=# OP ,r1=# OP1,r2=# OP2,r3=# OP3, thenr r1(or, for that matter,r r2orr r3) IS INTHE PLANE. Hence[(r2 r1) (r3 r1)] (r r1) = 0is a vector equation of the Normal vector a the equation of the plane that contains(1,0, 1),(3,1,4),(2, 2,0).Solution:We need 3 vectors. Pairing the points as shown yields (the order in which we subtract is irrelevant)(1,0, 1)&(3,1,4) u= 2i+j+ 5k,( )next(3,1,4)&(2, 2,0) v=i+ 3j+ 4k,( )and(2, 2,0)&(x, y, z) w= (x 2)i+ (y+ 2)j+zk.( )We also have that the cross productu v= 11i 3j+ 5kis a vector normal to the plane containing the given points.

5 Hence from the vector equation of theplanen (r r0) = 0we see that a vector equation of the plane is (u v) w= 0. This last equation implies 11(x 2) 3(y+ 2) + 5z= 0or 11x 3y+ 5z+ 16 = , we have the following definition. Two planesS1andS2with normal vectorsn1andn2,respectively, arei) perpendicular ifn1 n2= 0,ii) parallel ifn2=kn1, for some nonzero line of parametric equations for the line of intersection ofx y+ 2z= 1andx+y+z= 3 Solution:In a system of two equations and three unknowns, we choose one variable arbitrarily, sayz=t,and solve forxandyfromx y= 1 2t( )x+y= 3 t.( )Solving the system then givesx= 2 32t, y= 1 +12t, z= are the parametric equations for the lineLof intersection of the given Point of the point of intersection of the plane3x 2y+z= 5and the linex= 1+t, y= 2+2t, z= :If (x0, y0, z0) denotes the point of intersection, then we must have3x0 2y0+z0= 5, x0= 1 +t0, y0= 2 + 2t, z= 4t0,for some numbert0.

6 Substituting the latter equations into the equation of the plane gives3(1 +t0) 2( 2 + 2t0) + 4t0= 5ort0= the parametric equations for the lines, we then obtainx0= 3, y0= 10 andz0= 16. Sothe point is ( 3, 10, 16). Example DistanceDfrom a point to a (x0, y0, z0)be a point on the planeax+by+cz+d= 0and letnbe a normal vector tothe :IfP1(x1, y1, z1) is any point not on the plane, then the distanceDfrom a point to a plane is givenbyD=|ax1+by1+cz1+d| a2+b2+c2 Homework: prove this DistanceDfrom a point to a (2,1,4)be a point on the planex 3y+z 6 = 0. Find the distance fromP0to the :Do this as Angle between angle between two planes is defined to be the acute angle between their normal vectors.