246 VECTOR GEOMETRY
(b) Find the point on the line corresponding to
λ=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 )k
which 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 =3 anda 3 =−1 andb=i− 2 j+ 3 k, thenb 1 =1,b 2 =−2 andb 3 = 3
Hence, 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 equation2 x− 1
3=y+ 4
3=−z+ 5
2represents a straight line. Express this in vec-
tor 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 =−4 anda 3 =5 andb 1 =3
2,b 2 =3 andb 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)kor r=1
2( 1 + 3 λ)i+( 3 λ− 4 )j+( 5 − 2 λ)kNow try the following exercise.Exercise 99 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 k
which is parallel to the vector 2i+ 7 j− 4 k.
Determine the point on the line corresponding
toλ=2 in the resulting equation
[
r=(5+ 2 λ)i+(7λ−2)j
+(3− 4 λ)k;
r= 9 i+ 12 j− 5 k
]- Express the vector equation of the line in
problem 1 in standard Cartesian form.
[
x− 5
2
=y+ 2
7=3 −z
4=λ]In problems 3 and 4, express the given straight
line equations in vector form.3.3 x− 1
4=5 y+ 1
2=4 −z
3
[
r=^13 (1+ 4 λ)i+^15 (2λ−1)j
+(4− 3 λ)k]- 2x+ 1 =
1 − 4 y
5=3 z− 1
4
[
r=^12 (λ−1)i+^14 (1− 5 λ)j
+^13 (1+ 4 λ)k]