284 Higher Engineering Mathematics
magnitude, direction and sense), thenAP=λb,whereλ
is a scalar quantity. Hence, from above,r=a+λb (8)If, say,r=xi+yj+zk,a=a 1 i+a 2 j+a 3 kand
b=b 1 i+b 2 j+b 3 k, then from equation (8),xi+yj+zk=(a 1 i+a 2 j+a 3 k)+λ(b 1 i+b 2 j+b 3 k)Hence x=a 1 +λb 1 , y=a 2 +λb 2 and z=a 3 +λb 3.
Solving forλgives:x−a 1
b 1=y−a 2
b 2=z−a 3
b 3=λ (9)Equation(9)isthestandardCartesianformforthevector
equation of a straight line.Problem 11. (a) Determine the vector equation of
the line through the point with position vector
2 i+ 3 j−kwhich is parallel to the vectori− 2 j+ 3 k.
(b) Find the point on the line corresponding toλ= 3
in the resulting equation of part (a).
(c) Express the vector equation of the line in
standard Cartesian form.(a) From equation (8),r=a+λbi.e. r=( 2 i+ 3 j−k)+λ(i− 2 j+ 3 k)or r=( 2 +λ)i+( 3 − 2 λ)j+( 3 λ− 1 )kwhich is the vector equation of the line.
(b) Whenλ=3, r= 5 i− 3 j+ 8 k.
(c) From equation (9),
x−a 1
b 1=y−a 2
b 2=z−a 3
b 3=λSince a= 2 i+ 3 j−k,thena 1 = 2 ,a 2 =3anda 3 =−1andb=i− 2 j+ 3 k,thenb 1 = 1 ,b 2 =−2andb 3 = 3Hence, the Cartesian equations are:x− 2
1=y− 3
− 2=z−(− 1 )
3=λi.e. x− 2 =3 −y
2=z+ 1
3=λProblem 12. The equation
2 x− 1
3=y+ 4
3=−z+ 5
2
represents a straight line. Express this in vector
form.Comparing the given equation with equation (9), shows
that the coefficients ofx,yandzneed to be equal to
unity.Thus2 x− 1
3=y+ 4
3=−z+ 5
2becomes:x−^12
3
2=y+ 4
3=z− 5
− 2Again, comparing with equation (9), shows thata 1 =1
2
,a 2 =−4anda 3 =5andb 1 =3
2,b 2 =3andb 3 =− 2In vector form the equation is:r=(a 1 +λb 1 )i+(a 2 +λb 2 )j+(a 3 +λb 3 )k,
from equation (8)i.e.r=(
1
2+3
2λ)
i+(− 4 + 3 λ)j+( 5 − 2 λ)korr=1
2( 1 + 3 λ)i+( 3 λ− 4 )j+( 5 − 2 λ)kNow try the following exerciseExercise 114 Further problems on the
vector equation of a line- Find the vector equation of the line through the
point with position vector 5i− 2 j+ 3 kwhich