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+λb
i.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
=λ
Since a= 2 i+ 3 j−k,thena 1 = 2 ,
a 2 =3anda 3 =−1and
b=i− 2 j+ 3 k,then
b 1 = 1 ,b 2 =−2andb 3 = 3
Hence, the Cartesian equations are:
x− 2
1
=
y− 3
− 2
=
z−(− 1 )
3
=λ
i.e. x− 2 =
3 −y
2
=
z+ 1
3
=λ
Problem 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.
Thus
2 x− 1
3
=
y+ 4
3
=
−z+ 5
2
becomes:
x−^12
3
2
=
y+ 4
3
=
z− 5
− 2
Again, comparing with equation (9), shows that
a 1 =
1
2
,a 2 =−4anda 3 =5and
b 1 =
3
2
,b 2 =3andb 3 =− 2
In 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
2
λ
)
i+(− 4 + 3 λ)j+( 5 − 2 λ)k
orr=
1
2
( 1 + 3 λ)i+( 3 λ− 4 )j+( 5 − 2 λ)k
Now try the following exercise
Exercise 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