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Let
l
,
m
,
n
be the direction cosines of the line which is perpendicular to the line with direction cosines
l
1
,
m
1
,
n
1
and
l
2
,
m
2
,
n
2
.
l
,
m
,
n
are the direction cosines of the line.
∴
l
2
+
m
2
+
n
2
= 1 … (5)
It is known that,
Substituting the values from equations (5) and (6) in equation (4), we obtain
Thus, the direction cosines of the required line are
Thus, the angle between the lines is 90°
Question 4.Find the equation of the line parallel to x-axis and passing through the origin.
Solution :
The line parallel to
x
-axis and passing through the origin is
x
-axis itself.
Let A be a point on
x
-axis. Therefore, the coordinates of A are given by (
a
, 0, 0), where
a
∈ R.
Direction ratios of OA are (
a
− 0) =
a
, 0, 0
The equation of OA is given by,
Thus, the equation of line parallel to
x
-axis and passing through origin is
x/1 = y/0 = z/0
Question
5.If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Solution :
The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (−4, 3, −6), and
(2, 9, 2) respectively.
The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4
The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8
It can be seen that, a1/a2 = b1/b2 = c1/c2 = 1/2
Therefore, AB is parallel to CD.
Thus, the angle between AB and CD is either 0° or 180°.
Question
6.If the lines
are perpendicular, find the value of k
Solution :
The direction of ratios of the lines,
Therefore, for k= -10/7, the given lines are perpendicular to each other.
Question
7.Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane
Solution :
The position vector of the point (1, 2, 3) is
The direction ratios of the normal to the plane,
The equation of a line passing through a point and perpendicular to the given plane is given by,
Question
8.Find the equation of the plane passing through (
a
,
b
,
c
) and parallel to the plane
Solution :
Any plane parallel to the plane,
.......(1)
The plane passes through the point (
a
,
b
,
c
). Therefore, the position vector
.png)
of this point is
Therefore, equation (1) becomes
Substituting
in equation (1), we obtain
This is the vector equation of the required plane.
Substituting
in equation (2), we obtain
Question
9.Find the shortest distance between lines
Solution :
The given lines are
It is known that the shortest distance between two lines,
is given by
.png)
Comparing
Substituting all the values in equation (1), we obtain
Therefore, the shortest distance between the two given lines is 9 units.
Question
10.Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
Solution :
It is known that the equation of the line passing through the points, (
x
1
,
y
1
,
z
1
) and (
x
2
,
y
2
,
z
2
), is
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,
Any point on the line is of the form (5 − 2
k
, 3
k
+ 1, 6 −5
k
).
The equation of YZ-plane is
x
= 0
Since the line passes through YZ-plane,
5 − 2
k
= 0
Therefore, the required point is
.
.png)
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,
Any point on the line is of the form (5 − 2
k
, 3
k
+ 1, 6 −5
k
).
Since the line passes through ZX-plane,
Question
12
. Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1) crosses the plane 2
x
+
y
+
z
= 7).
Solution :
It is known that the equation of the line through the points, (
x
1
,
y
1
,
z
1
) and (
x
2
,
y
2
,
z
2
), is
Therefore, any point on the line is of the form (3 −
k
,
k
− 4, 6
k
− 5).
This point lies on the plane, 2
x
+
y
+
z
= 7
∴ 2 (3 −
k
) + (
k
− 4) + (6
k
− 5) = 7
5k - 3 = 7
k = 2
Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 × 2 − 5) i.e.,
(1, −2, 7).
Question
13.
Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes
x
+ 2
y
+ 3
z
= 5 and 3
x
+ 3
y
+
z
= 0.
Solution :
The equation of the plane passing through the point (−1, 3, 2) is
a
(
x
+ 1) +
b
(
y
− 3) +
c
(
z
− 2) = 0 … (1)
where,
a
,
b
,
c
are the direction ratios of normal to the plane.
It is known that two planes,
Plane (1) is perpendicular to the plane,
x
+ 2
y
+ 3
z
= 5
a.1 + b .2 + c.3 = 0
a + 2b + 3c = 0 ....(2)
Also, plane (1) is perpendicular to the plane, 3
x
+ 3
y
+
z
= 0
a .3 + b.3 + c.1 = 0
3a + 3b + c = 0 .....(3)
From equations (2) and (3), we obtain
Substituting the values of
a
,
b
, and
c
in equation (1), we obtain
This is the required equation of the plane.
Question
14. If the points (1, 1,
p
) and (−3, 0, 1) be equidistant from the plane
then find the value of p
Solution :
The position vector through the point (1, 1,
p
) is
Similarly, the position vector through the point (−3, 0, 1) is
The equation of the given plane is
.png)
and the plane,
is given by,
Here,
and
d
= -13
Therefore, the distance between the point (1, 1,
p
) and the given plane is
Similarly, the distance between the point (−3, 0, 1) and the given plane is
It is given that the distance between the required plane and the points, (1, 1,
p
) and (−3, 0, 1), is equal.
∴
D
1
=
D
2
Question
15.Find the equation of the plane passing through the line of intersection of the planes
and
and parallel to x - axis.
Solution :
Equation of one plane is
The equation of any plane passing through the line of intersection of these planes is
Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).
The required plane is parallel to
x
-axis. Therefore, its normal is perpendicular to
x
-axis.
The direction ratios of
x
-axis are 1, 0, and 0.
Substituting λ = -1/2 in equation (1), we obtain
Therefore, its Cartesian equation is
y
− 3
z
+ 6 = 0
This is the equation of the required plane.
and which is perpendicular to the plane
Solution :
The equations of the given planes are
The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is
The plane in equation (3) is perpendicular to the plane,
Substituting λ = 7/19 in equation (3), we obtain
This is the vector equation of the required plane.
The Cartesian equation of this plane can be obtained by substituting
in equation (3).
Question
18.Find the distance of the point (−1, −5, −10) from the point of intersection of the line
and the plane
Solution :
The equation of the given line is
Substituting this value in equation (1), we obtain the equation of the line as
This means that the position vector of the point of intersection of the line and the plane is
Question
19.Find the vector equation of the line passing through (1, 2, 3) and parallel to the plane
and
Solution :
Let the required line be parallel to vector
.png)
given by,
The position vector of the point (1, 2, 3) is
The equation of line passing through (1, 2, 3) and parallel to
The equations of the given planes are
From equations (4) and (5), we obtain
Therefore, the direction ratios of
Substituting the value of
This is the equation of the required line.
Question
20.Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines:
Solution :
Let the required line be parallel to the vector
The equation of the line passing through (1, 2, −4) and parallel to vector
The equations of the lines are
Line (1) and line (2) are perpendicular to each other.
Also, line (1) and line (3) are perpendicular to each other.
From equations (4) and (5), we obtain
Direction ratios of
Substituting
This is the equation of the required line.
Question
21.Prove that if a plane has the intercepts a,b,c and is at a distance of P units from the origin, then
Solution :
The equation of a plane having intercepts
a
,
b
,
c
with
x
,
y
, and
z
axes respectively is given by,
The distance (
p
) of the plane from the origin is given by,
Thus, the distance between the lines is 2/√29 units.
Hence, the correct answer is D.
Question
23.
The planes: 2
x
−
y
+ 4
z
= 5 and 5
x
− 2.5
y
+ 10
z
= 6 are
Therefore, the given planes are parallel.
Hence, the correct answer is B.
