284 Higher Engineering Mathematics
magnitude, direction and sense), then AP = λb, where λ
is a scalar quantity. Hence, from above,
r = a +λ b
(8)
If, say, r = xi + yj + zk, a =a 1 i +a 2 j + a 3 k and
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 for λ gives:
x −a 1
b 1
=
y − a 2
b 2
=
z − a 3
b 3
= λ
(9)
Equation (9) is the standard Cartesian form for the vector
equation of a straight line.
Problem 11. (a) Determine the vector equation of
the line through the point with position vector
2i + 3j −k which is parallel to the vector i − 2j + 3k.
(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 + λb
i.e. r = (2i + 3j −k) +λ(i − 2j + 3k)
or r = (2 + λ)i + (3 − 2λ)j + (3λ − 1)k
which is the vector equation of the line.
(b) When λ =3, r = 5i −3j + 8k.
(c) From equation (9),
x − a 1
b 1
=
y − a 2
b 2
=
z − a 3
b 3
= λ
Since a = 2i + 3j − k, then a 1 = 2,
a 2 = 3 and a 3 = −1 and
b = i − 2j + 3k, then
b 1 = 1, b 2 = −2 and b 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
2x − 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 of x, y and z need to be equal to
unity.
Thus
2x − 1
3
=
y + 4
3
=
−z + 5
2
becomes:
x −
1
2
3
2
=
y + 4
3
=
z − 5
−2
Again, comparing with equation (9), shows that
a 1 =
1
2
, a 2 = −4 and a 3 = 5 and
b 1 =
3
2
, b 2 = 3 and b 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 + 3λ) j + (5 − 2λ)k
or r =
1
2
(1 + 3λ)i + (3λ − 4) j + (5 − 2λ)k
Now try the following exercise
Exercise 114 Further problems on the
vector equation of a line
1. Find the vector equation of the line through the
point with position vector 5i −2j + 3k which
magnitude, direction and sense), then AP = λb, where λ
is a scalar quantity. Hence, from above,
r = a +λ b
(8)
If, say, r = xi + yj + zk, a =a 1 i +a 2 j + a 3 k and
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 for λ gives:
x −a 1
b 1
=
y − a 2
b 2
=
z − a 3
b 3
= λ
(9)
Equation (9) is the standard Cartesian form for the vector
equation of a straight line.
Problem 11. (a) Determine the vector equation of
the line through the point with position vector
2i + 3j −k which is parallel to the vector i − 2j + 3k.
(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 + λb
i.e. r = (2i + 3j −k) +λ(i − 2j + 3k)
or r = (2 + λ)i + (3 − 2λ)j + (3λ − 1)k
which is the vector equation of the line.
(b) When λ =3, r = 5i −3j + 8k.
(c) From equation (9),
x − a 1
b 1
=
y − a 2
b 2
=
z − a 3
b 3
= λ
Since a = 2i + 3j − k, then a 1 = 2,
a 2 = 3 and a 3 = −1 and
b = i − 2j + 3k, then
b 1 = 1, b 2 = −2 and b 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
2x − 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 of x, y and z need to be equal to
unity.
Thus
2x − 1
3
=
y + 4
3
=
−z + 5
2
becomes:
x −
1
2
3
2
=
y + 4
3
=
z − 5
−2
Again, comparing with equation (9), shows that
a 1 =
1
2
, a 2 = −4 and a 3 = 5 and
b 1 =
3
2
, b 2 = 3 and b 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 + 3λ) j + (5 − 2λ)k
or r =
1
2
(1 + 3λ)i + (3λ − 4) j + (5 − 2λ)k
Now try the following exercise
Exercise 114 Further problems on the
vector equation of a line
1. Find the vector equation of the line through the
point with position vector 5i −2j + 3k which