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Three Dimensional Geometry
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NCERT Solutions
Three Dimensional Geometry
25 Solutions
Exercise:
All Exercises
EXERCISE 11.1
EXERCISE 11.2
Miscellaneous Exercise on Chapter 11
Q1
EXERCISE 11.1
If a line makes angles
90
∘
,
135
∘
,
45
∘
90^{\circ}, 135^{\circ}, 45^{\circ}
9
0
∘
,
13
5
∘
,
4
5
∘
with the
x
,
y
x, y
x
,
y
and
z
z
z
-axes respectively, find its direction cosines.
Q2
EXERCISE 11.1
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Q3
EXERCISE 11.1
If a line has the direction ratios
−
18
,
12
,
−
4
-18, 12, -4
−
18
,
12
,
−
4
, then what are its direction cosines ?
Q4
EXERCISE 11.1
Show that the points
(
2
,
3
,
4
)
,
(
−
1
,
−
2
,
1
)
,
(
5
,
8
,
7
)
(2, 3, 4), (-1, -2, 1), (5, 8, 7)
(
2
,
3
,
4
)
,
(
−
1
,
−
2
,
1
)
,
(
5
,
8
,
7
)
are collinear.
Q5
EXERCISE 11.1
Find the direction cosines of the sides of the triangle whose vertices are
(
3
,
5
,
−
4
)
,
(
−
1
,
1
,
2
)
(3, 5, -4), (-1, 1, 2)
(
3
,
5
,
−
4
)
,
(
−
1
,
1
,
2
)
and
(
−
5
,
−
5
,
−
2
)
(-5, -5, -2)
(
−
5
,
−
5
,
−
2
)
.
Q1
EXERCISE 11.2
Show that the three lines with direction cosines
12
13
,
−
3
13
,
−
4
13
;
4
13
,
12
13
,
3
13
;
3
13
,
−
4
13
,
12
13
\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}
13
12
,
13
−
3
,
13
−
4
;
13
4
,
13
12
,
13
3
;
13
3
,
13
−
4
,
13
12
are mutually perpendicular.
Q2
EXERCISE 11.2
Show that the line through the points
(
1
,
−
1
,
2
)
,
(
3
,
4
,
−
2
)
(1, -1, 2), (3, 4, -2)
(
1
,
−
1
,
2
)
,
(
3
,
4
,
−
2
)
is perpendicular to the line through the points
(
0
,
3
,
2
)
(0, 3, 2)
(
0
,
3
,
2
)
and
(
3
,
5
,
6
)
(3, 5, 6)
(
3
,
5
,
6
)
.
Q3
EXERCISE 11.2
Show that the line through the points
(
4
,
7
,
8
)
,
(
2
,
3
,
4
)
(4, 7, 8), (2, 3, 4)
(
4
,
7
,
8
)
,
(
2
,
3
,
4
)
is parallel to the line through the points
(
−
1
,
−
2
,
1
)
,
(
1
,
2
,
5
)
(-1, -2, 1), (1, 2, 5)
(
−
1
,
−
2
,
1
)
,
(
1
,
2
,
5
)
.
Q4
EXERCISE 11.2
Find the equation of the line which passes through the point
(
1
,
2
,
3
)
(1, 2, 3)
(
1
,
2
,
3
)
and is parallel to the vector
3
i
^
+
2
j
^
−
2
k
^
3 \hat{i} + 2 \hat{j} - 2 \hat{k}
3
i
^
+
2
j
^
−
2
k
^
.
Q5
EXERCISE 11.2
Find the equation of the line in vector and in cartesian form that passes through the point with position vector
2
i
^
−
j
+
4
k
^
2 \hat{i} - j + 4 \hat{k}
2
i
^
−
j
+
4
k
^
and is in the direction
i
^
+
2
j
^
−
k
^
\hat{i} + 2 \hat{j} - \hat{k}
i
^
+
2
j
^
−
k
^
.
Q6
EXERCISE 11.2
Find the cartesian equation of the line which passes through the point ( -2, 4, -5 ) and parallel to the line given by
x
+
3
3
=
y
−
4
5
=
z
+
8
6
\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}
3
x
+
3
=
5
y
−
4
=
6
z
+
8
.
Q7
EXERCISE 11.2
The cartesian equation of a line is
x
−
5
3
=
y
+
4
7
=
z
−
6
2
\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}
3
x
−
5
=
7
y
+
4
=
2
z
−
6
. Write its vector form.
Q8
EXERCISE 11.2
Find the angle between the following pairs of lines:
(i)
r
⃗
=
2
i
^
−
5
j
^
+
k
^
+
λ
(
3
i
^
+
2
j
^
+
6
k
^
)
\vec{r} = 2 \hat{i} - 5 \hat{j} + \hat{k} + \lambda(3 \hat{i} + 2 \hat{j} + 6 \hat{k})
r
=
2
i
^
−
5
j
^
+
k
^
+
λ
(
3
i
^
+
2
j
^
+
6
k
^
)
and
r
⃗
=
7
i
^
−
6
k
^
+
μ
(
i
^
+
2
j
^
+
2
k
^
)
\vec{r} = 7 \hat{i} - 6 \hat{k} + \mu(\hat{i} + 2 \hat{j} + 2 \hat{k})
r
=
7
i
^
−
6
k
^
+
μ
(
i
^
+
2
j
^
+
2
k
^
)
(ii)
r
⃗
=
3
i
^
+
j
^
−
2
k
^
+
λ
(
i
^
−
j
^
−
2
k
^
)
\vec{r} = 3 \hat{i} + \hat{j} - 2 \hat{k} + \lambda(\hat{i} - \hat{j} - 2 \hat{k})
r
=
3
i
^
+
j
^
−
2
k
^
+
λ
(
i
^
−
j
^
−
2
k
^
)
and
r
⃗
=
2
i
^
−
j
^
−
56
k
^
+
μ
(
3
i
^
−
5
j
^
−
4
k
^
)
\vec{r} = 2 \hat{i} - \hat{j} - 56 \hat{k} + \mu(3 \hat{i} - 5 \hat{j} - 4 \hat{k})
r
=
2
i
^
−
j
^
−
56
k
^
+
μ
(
3
i
^
−
5
j
^
−
4
k
^
)
Q9
EXERCISE 11.2
Find the angle between the following pair of lines:
(i)
x
−
2
2
=
y
−
1
5
=
z
+
3
−
3
\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}
2
x
−
2
=
5
y
−
1
=
−
3
z
+
3
and
x
+
2
−
1
=
y
−
4
8
=
z
−
5
4
\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}
−
1
x
+
2
=
8
y
−
4
=
4
z
−
5
(ii)
x
2
=
y
2
=
z
1
\frac{x}{2} = \frac{y}{2} = \frac{z}{1}
2
x
=
2
y
=
1
z
and
x
−
5
4
=
y
−
2
1
=
z
−
3
8
\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}
4
x
−
5
=
1
y
−
2
=
8
z
−
3
Q10
EXERCISE 11.2
Find the values of
p
p
p
so that the lines
1
−
x
3
=
7
y
−
14
2
p
=
z
−
3
2
\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}
3
1
−
x
=
2
p
7
y
−
14
=
2
z
−
3
and
7
−
7
x
3
p
=
y
−
5
1
=
6
−
z
5
\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}
3
p
7
−
7
x
=
1
y
−
5
=
5
6
−
z
are at right angles.
Q11
EXERCISE 11.2
Show that the lines
x
−
5
7
=
y
+
2
−
5
=
z
1
\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}
7
x
−
5
=
−
5
y
+
2
=
1
z
and
x
1
=
y
2
=
z
3
\frac{x}{1} = \frac{y}{2} = \frac{z}{3}
1
x
=
2
y
=
3
z
are perpendicular to each other.
Q12
EXERCISE 11.2
Find the shortest distance between the lines
r
⃗
=
(
i
^
+
2
j
^
+
k
^
)
+
λ
(
i
^
−
j
^
+
k
^
)
\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})
r
=
(
i
^
+
2
j
^
+
k
^
)
+
λ
(
i
^
−
j
^
+
k
^
)
and
r
⃗
=
2
i
^
−
j
^
−
k
^
+
μ
(
2
i
^
+
j
^
+
2
k
^
)
\vec{r} = 2\hat{i} - \hat{j} - \hat{k} + \mu(2\hat{i} + \hat{j} + 2\hat{k})
r
=
2
i
^
−
j
^
−
k
^
+
μ
(
2
i
^
+
j
^
+
2
k
^
)
Q13
EXERCISE 11.2
Find the shortest distance between the lines
x
+
1
7
=
y
+
1
−
6
=
z
+
1
1
\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}
7
x
+
1
=
−
6
y
+
1
=
1
z
+
1
and
x
−
3
1
=
y
−
5
−
2
=
z
−
7
1
\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}
1
x
−
3
=
−
2
y
−
5
=
1
z
−
7
Q14
EXERCISE 11.2
Find the shortest distance between the lines whose vector equations are
r
⃗
=
(
i
^
+
2
j
^
+
3
k
^
)
+
λ
(
i
^
−
3
j
^
+
2
k
^
)
\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})
r
=
(
i
^
+
2
j
^
+
3
k
^
)
+
λ
(
i
^
−
3
j
^
+
2
k
^
)
and
r
⃗
=
4
i
^
+
5
j
^
+
6
k
^
+
μ
(
2
i
^
+
3
j
^
+
k
^
)
\vec{r} = 4\hat{i} + 5\hat{j} + 6\hat{k} + \mu(2\hat{i} + 3\hat{j} + \hat{k})
r
=
4
i
^
+
5
j
^
+
6
k
^
+
μ
(
2
i
^
+
3
j
^
+
k
^
)
Q15
EXERCISE 11.2
Find the shortest distance between the lines whose vector equations are
r
⃗
=
(
1
−
t
)
i
^
+
(
t
−
2
)
j
^
+
(
3
−
2
t
)
k
^
\vec{r} = (1-t) \hat{i} + (t-2) \hat{j} + (3-2t) \hat{k}
r
=
(
1
−
t
)
i
^
+
(
t
−
2
)
j
^
+
(
3
−
2
t
)
k
^
and
r
⃗
=
(
s
+
1
)
i
^
+
(
2
s
−
1
)
j
^
−
(
2
s
+
1
)
k
^
\vec{r} = (s+1) \hat{i} + (2s-1) \hat{j} - (2s+1) \hat{k}
r
=
(
s
+
1
)
i
^
+
(
2
s
−
1
)
j
^
−
(
2
s
+
1
)
k
^
Q1
Miscellaneous Exercise on Chapter 11
Find the angle between the lines whose direction ratios are
a
,
b
,
c
a, b, c
a
,
b
,
c
and
b
−
c
,
c
−
a
,
a
−
b
b-c, c-a, a-b
b
−
c
,
c
−
a
,
a
−
b
.
Q2
Miscellaneous Exercise on Chapter 11
Find the equation of a line parallel to
x
x
x
-axis and passing through the origin.
Q3
Miscellaneous Exercise on Chapter 11
If the lines
x
−
1
−
3
=
y
−
2
2
k
=
z
−
3
2
\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}
−
3
x
−
1
=
2
k
y
−
2
=
2
z
−
3
and
x
−
1
3
k
=
y
−
1
1
=
z
−
6
−
5
\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}
3
k
x
−
1
=
1
y
−
1
=
−
5
z
−
6
are perpendicular, find the value of
k
k
k
.
Q4
Miscellaneous Exercise on Chapter 11
Find the shortest distance between lines
r
⃗
=
6
i
^
+
2
j
^
+
2
k
^
+
λ
(
i
^
−
2
j
^
+
2
k
^
)
\vec{r} = 6\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})
r
=
6
i
^
+
2
j
^
+
2
k
^
+
λ
(
i
^
−
2
j
^
+
2
k
^
)
and
r
⃗
=
−
4
i
^
−
k
^
+
μ
(
3
i
^
−
2
j
^
−
2
k
^
)
\vec{r} = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})
r
=
−
4
i
^
−
k
^
+
μ
(
3
i
^
−
2
j
^
−
2
k
^
)
.
Q5
Miscellaneous Exercise on Chapter 11
Find the vector equation of the line passing through the point
(
1
,
2
,
−
4
)
(1, 2, -4)
(
1
,
2
,
−
4
)
and perpendicular to the two lines:
x
−
8
3
=
y
+
19
−
16
=
z
−
10
7
\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}
3
x
−
8
=
−
16
y
+
19
=
7
z
−
10
and
x
−
15
3
=
y
−
29
8
=
z
−
5
−
5
\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}
3
x
−
15
=
8
y
−
29
=
−
5
z
−
5
.
More from this chapter
Chapter overview
Important Points
Practice Questions
Flashcards