NCERT Solutions: Exercise - 11.2 - Three Dimensional Geometry

# Exercise - 11.2 - Three Dimensional Geometry NCERT Solutions | Mathematics (Maths) Class 12 - JEE PDF Download

``` Page 1

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
EXERCISE 11.2                                                                        PAGE NO: 477
1. Show that the three lines with direction cosines
Are mutually perpendicular.
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 ? is the acute angle between the two lines,
Then cos ? = |l 1l 2 + m 1m 2 + n 1n 2|
If two lines are perpendicular, then the angle between the two is ? = 90°
For perpendicular lines, | l 1l 2 + m 1m 2 + n 1n 2 | = cos 90° = 0, i.e. | l 1l 2 + m 1m 2 + n 1n 2 | = 0
So, in order to check if the three lines are mutually perpendicular, we compute | l 1l 2 + m 1m 2 + n 1n 2 | for all the pairs of
the three lines.
Firstly let us compute, | l 1l 2 + m 1m 2 + n 1n 2 |

So,  L 1 ? L 2 …… (1)
Similarly,
Let us compute, | l 2l 3 + m 2m 3 + n 2n 3 |

Page 2

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
EXERCISE 11.2                                                                        PAGE NO: 477
1. Show that the three lines with direction cosines
Are mutually perpendicular.
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 ? is the acute angle between the two lines,
Then cos ? = |l 1l 2 + m 1m 2 + n 1n 2|
If two lines are perpendicular, then the angle between the two is ? = 90°
For perpendicular lines, | l 1l 2 + m 1m 2 + n 1n 2 | = cos 90° = 0, i.e. | l 1l 2 + m 1m 2 + n 1n 2 | = 0
So, in order to check if the three lines are mutually perpendicular, we compute | l 1l 2 + m 1m 2 + n 1n 2 | for all the pairs of
the three lines.
Firstly let us compute, | l 1l 2 + m 1m 2 + n 1n 2 |

So,  L 1 ? L 2 …… (1)
Similarly,
Let us compute, | l 2l 3 + m 2m 3 + n 2n 3 |

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry

So, L 2 ? L 3 ….. (2)
Similarly,
Let us compute, | l 3l 1 + m 3m 1 + n 3n 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 1a 2 + b 1b 2 + c 1c 2 = 0
a 1a 2 + b 1b 2 + c 1c 2 = 2(3) + 5(2) + 4(-4)
Page 3

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
EXERCISE 11.2                                                                        PAGE NO: 477
1. Show that the three lines with direction cosines
Are mutually perpendicular.
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 ? is the acute angle between the two lines,
Then cos ? = |l 1l 2 + m 1m 2 + n 1n 2|
If two lines are perpendicular, then the angle between the two is ? = 90°
For perpendicular lines, | l 1l 2 + m 1m 2 + n 1n 2 | = cos 90° = 0, i.e. | l 1l 2 + m 1m 2 + n 1n 2 | = 0
So, in order to check if the three lines are mutually perpendicular, we compute | l 1l 2 + m 1m 2 + n 1n 2 | for all the pairs of
the three lines.
Firstly let us compute, | l 1l 2 + m 1m 2 + n 1n 2 |

So,  L 1 ? L 2 …… (1)
Similarly,
Let us compute, | l 2l 3 + m 2m 3 + n 2n 3 |

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry

So, L 2 ? L 3 ….. (2)
Similarly,
Let us compute, | l 3l 1 + m 3m 1 + n 3n 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 1a 2 + b 1b 2 + c 1c 2 = 0
a 1a 2 + b 1b 2 + c 1c 2 = 2(3) + 5(2) + 4(-4)

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
= 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 .
Page 4

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
EXERCISE 11.2                                                                        PAGE NO: 477
1. Show that the three lines with direction cosines
Are mutually perpendicular.
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 ? is the acute angle between the two lines,
Then cos ? = |l 1l 2 + m 1m 2 + n 1n 2|
If two lines are perpendicular, then the angle between the two is ? = 90°
For perpendicular lines, | l 1l 2 + m 1m 2 + n 1n 2 | = cos 90° = 0, i.e. | l 1l 2 + m 1m 2 + n 1n 2 | = 0
So, in order to check if the three lines are mutually perpendicular, we compute | l 1l 2 + m 1m 2 + n 1n 2 | for all the pairs of
the three lines.
Firstly let us compute, | l 1l 2 + m 1m 2 + n 1n 2 |

So,  L 1 ? L 2 …… (1)
Similarly,
Let us compute, | l 2l 3 + m 2m 3 + n 2n 3 |

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry

So, L 2 ? L 3 ….. (2)
Similarly,
Let us compute, | l 3l 1 + m 3m 1 + n 3n 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 1a 2 + b 1b 2 + c 1c 2 = 0
a 1a 2 + b 1b 2 + c 1c 2 = 2(3) + 5(2) + 4(-4)

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
= 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 .

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
Solution:

5. Find the equation of the line in vector and in Cartesian form that passes through the point with position
vector and  is in the direction
Solution:
Page 5

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
EXERCISE 11.2                                                                        PAGE NO: 477
1. Show that the three lines with direction cosines
Are mutually perpendicular.
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 ? is the acute angle between the two lines,
Then cos ? = |l 1l 2 + m 1m 2 + n 1n 2|
If two lines are perpendicular, then the angle between the two is ? = 90°
For perpendicular lines, | l 1l 2 + m 1m 2 + n 1n 2 | = cos 90° = 0, i.e. | l 1l 2 + m 1m 2 + n 1n 2 | = 0
So, in order to check if the three lines are mutually perpendicular, we compute | l 1l 2 + m 1m 2 + n 1n 2 | for all the pairs of
the three lines.
Firstly let us compute, | l 1l 2 + m 1m 2 + n 1n 2 |

So,  L 1 ? L 2 …… (1)
Similarly,
Let us compute, | l 2l 3 + m 2m 3 + n 2n 3 |

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry

So, L 2 ? L 3 ….. (2)
Similarly,
Let us compute, | l 3l 1 + m 3m 1 + n 3n 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 1a 2 + b 1b 2 + c 1c 2 = 0
a 1a 2 + b 1b 2 + c 1c 2 = 2(3) + 5(2) + 4(-4)

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
= 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 .

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry
Solution:

5. Find the equation of the line in vector and in Cartesian form that passes through the point with position
vector and  is in the direction
Solution:

NCERT Solutions for Class 12 Maths Chapter 11 –
Three Dimensional Geometry

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)
```

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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