

∴ The direction cosines are
-18/22, 12/22, -4/22 => -9/11, 6/11, -2/11
4. Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.
Solution:
If the direction ratios of two lines segments are proportional, then the lines are collinear.
Given:
A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7)
Direction ratio of line joining A (2, 3, 4) and B (−1, −2, 1), are
(−1−2), (−2−3), (1−4) = (−3, −5, −3)
Where, a
1
= -3, b
1
= -5, c
1
= -3
Direction ratio of line joining B (−1, −2, 1) and C (5, 8, 7) are
(5− (−1)), (8− (−2)), (7−1) = (6, 10, 6)
Where, a
2
= 6, b
2
= 10 and c
2
=6
Now,
∴ A, B, C are collinear.
5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (-1, 1, 2) and (–5, –5, –2).
Solution:
Given:
The vertices are (3, 5, –4), (-1, 1, 2) and (–5, –5, –2).
The direction cosines of the two points passing through A(x
1
, y
1
, z
1
) and B(x
2
, y
2
, z
2
) is given by (x
2
– x
1
), (y
2
-y
1
), (z
2
-z
1
)
Firstly let us find the direction ratios of AB
Where, A = (3, 5, -4) and B = (-1, 1, 2)
Ratio of AB = [(x
2
– x
1
)
2
, (y
2
– y
1
)
2
, (z
2
– z
1
)
2
]
= (-1-3), (1-5), (2-(-4)) = -4, -4, 6
Then by using the formula,
√[(x
2
– x
1
)
2
+ (y
2
– y
1
)
2
+ (z
2
– z
1
)
2
]
√[(-4)
2
+ (-4)
2
+ (6)
2
] = √(16+16+36)
= √68
= 2√17
Now let us find the direction cosines of the line AB
By using the formula,
-4/2√17 , -4/2√17, 6/2√17
Or -2/√17, -2/√17, 3/√17
Similarly,
Let us find the direction ratios of BC
Where, B = (-1, 1, 2) and C = (-5, -5, -2)
Ratio of AB = [(x
2
– x
1
)
2
, (y
2
– y
1
)
2
, (z
2
– z
1
)
2
]
= (-5+1), (-5-1), (-2-2) = -4, -6, -4
Then by using the formula,
√[(x
2
– x
1
)
2
+ (y
2
– y
1
)
2
+ (z
2
– z
1
)
2
]
√[(-4)
2
+ (-6)
2
+ (-4)
2
] = √(16+36+16)
= √68
= 2√17
Now, let us find the direction cosines of the line AB
By using the formula,
-4/2√17, -6/2√17, -4/2√17
Or -2/√17, -3/√17, -2/√17
Similarly,
Let us find the direction ratios of CA
Where, C = (-5, -5, -2) and A = (3, 5, -4)
Ratio of AB = [(x
2
– x
1
)
2
, (y
2
– y
1
)
2
, (z
2
– z
1
)
2
]
= (3+5), (5+5), (-4+2) = 8, 10, -2
Then, by using the formula,
√[(x
2
– x
1
)
2
+ (y
2
– y
1
)
2
+ (z
2
– z
1
)
2
]
√[(8)
2
+ (10)
2
+ (-2)
2
] = √(64+100+4)
= √168
= 2√42
Now, let us find the direction cosines of the line AB
By using the formula,
8/2√42, 10/2√42, -2/2√42
Or 4/√42, 5/√42, -1/√42
Solution:
Let us consider the direction cosines of L
1
, L
2
and L
3
be l
1
, m
1
, n
1
; l
2
, m
2
, n
2
and l
3
, m
3
, n
3
.
We know that
If l
1
, m
1
, n
1
and l
2
, m
2
, n
2
are the direction cosines of two lines,
And θ is the acute angle between the two lines,
Then cos θ = |l
1
l
2
+ m
1
m
2
+ n
1
n
2
|
If two lines are perpendicular, then the angle between the two is θ = 90°
For perpendicular lines, | l
1
l
2
+ m
1
m
2
+ n
1
n
2
| = cos 90° = 0, i.e. | l
1
l
2
+ m
1
m
2
+ n
1
n
2
| = 0
So, in order to check if the three lines are mutually perpendicular, we compute | l
1
l
2
+ m
1
m
2
+ n
1
n
2
| for all the pairs of the three lines.
Firstly let us compute, | l
1
l
2
+ m
1
m
2
+ n
1
n
2
|
So, L
1
⊥
L
2
…… (1)
Similarly,
Let us compute, | l
2
l
3
+ m
2
m
3
+ n
2
n
3
|
So, L
2
⊥
L
3
….. (2)
Similarly,
Let us compute, | l
3
l
1
+ m
3
m
1
+ n
3
n
1
|
So, L
1
⊥
L
3
….. (3)
∴ By (1), (2) and (3), the lines are perpendicular.
L
1
, L
2
and L
3
are mutually perpendicular.
2. Show that the line through the points (1, –1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Solution:
Given:
The points (1, –1, 2), (3, 4, –2) and (0, 3, 2), (3, 5, 6).
Let us consider AB be the line joining the points, (1, -1, 2) and (3, 4, -2), and CD be the line through the points (0, 3, 2) and (3, 5, 6).
Now,
The direction ratios, a
1
, b
1
, c
1
of AB are
(3 – 1), (4 – (-1)), (-2 – 2) = 2, 5, -4.
Similarly,
The direction ratios, a
2
, b
2
, c
2
of CD are
(3 – 0), (5 – 3), (6 – 2) = 3, 2, 4.
Then, AB and CD will be perpendicular to each other, if a
1
a
2
+ b
1
b
2
+ c
1
c
2
= 0
a
1
a
2
+ b
1
b
2
+ c
1
c
2
= 2(3) + 5(2) + 4(-4)
= 6 + 10 – 16
= 0
∴ AB and CD are perpendicular to each other.
3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).
Solution:
Given:
The points (4, 7, 8), (2, 3, 4) and (–1, –2, 1), (1, 2, 5).
Let us consider AB to be the line joining the points, (4, 7, 8), (2, 3, 4) and CD to be the line through the points (–1, –2, 1), (1, 2, 5).
Now,
The direction ratios, a
1
, b
1
, c
1
of AB are
(2 – 4), (3 – 7), (4 – 8) = -2, -4, -4.
The direction ratios, a
2
, b
2
, c
2
of CD are
(1 – (-1)), (2 – (-2)), (5 – 1) = 2, 4, 4.
Then, AB will be parallel to CD, if
So, a
1
/a
2
= -2/2 = -1
b
1
/b
2
= -4/4 = -1
c
1
/c
2
= -4/4 = -1
∴ We can say that,
-1 = -1 = -1
Hence, AB is parallel to CD where the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5)
4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector
Solution:
5. Find the equation of the line in vector and in Cartesian form that passes through the point with
position vector
is in the direction
Solution:
6. Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and parallel to the line given by
Solution:
Given:
The points (-2, 4, -5)
We know that the Cartesian equation of a line through a point (x
1
, y
1
, z
1
) and having direction ratios a, b, c is
7. The Cartesian equation of a line is
Write its vector form.
Solution:
So when comparing this standard form with the given equation, we get
x
1
= 5, y
1
= -4, z
1
= 6 and
l = 3, m = 7, n = 2
8. Find the vector and the Cartesian equations of the lines that passes through the origin and (5, –2, 3).
Solution:
9. Find the vector and the Cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).
Solution:
10.
Find the angle between the following pairs of lines:
Solution:
So,
By (3), we have
11. Find the angle between the following pair of lines:
Solution:
11.1 Introduction
11.2 Direction Cosines and Direction Ratios of a Line
11.2.1 Relation between the direction cosines of a line
11.2.2 Direction cosines of a line passing through two points
11.3 Equation of a Line in Space
11.3.1 Equation of a line through a given point and parallel to a given vector b.
11.3.2 Equation of a line passing through two given points
11.4 Angle between Two Lines
11.5 Shortest Distance between Two Lines
11.5.1 Distance between two skew lines
11.5.2 Distance between parallel lines
11.6 Plane
11.6.1 Equation of a plane in normal form
11.6.2 Equation of a plane perpendicular to a given vector, passing through a given point
11.6.3 Equation of a plane passing through three non-collinear points
11.6.4 Intercept form of the equation of a plane
11.6.5 Plane passing through the intersection of two given planes
11.7 Coplanarity of Two Lines
11.8 Angle between Two Planes
11.9 Distance of a Point from a Plane
11.10 Angle between a Line and a Plane
